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NCERT Exemplar Class 10 Maths Solutions Chapter 4 Quadratic Equations

NCERT Exemplar Class 10 Maths Solutions Chapter 4 Quadratic Equations

Edited By Safeer PP | Updated on Sep 05, 2022 01:22 PM IST | #CBSE Class 10th
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NCERT exemplar Class 10 Maths solutions chapter 4 provides the understanding for forming quadratic equations by the given set of information in the problems. These NCERT exemplar Class 10 Maths chapter 4 solutions are created by our experienced faculties of Mathematics and have undergone a strict quality check so that the students can get accurate and relevant solutions while reviewing the questions of NCERT Solutions for Class 10 Maths.

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These NCERT exemplar Class 10 Maths chapter 4 solutions due to their comprehensive nature provide useful insights to the students regarding the concepts of Quadratic Equations given in the NCERT. These NCERT exemplar Class 10 Maths solutions chapter 4 follow the prescribed CBSE syllabus for Class 10.

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Question:1

State whether the following quadratic equations have two distinct real roots. Justify your answer.
(i)x23x+4=0
(ii)2x2+x1=0
(iii)2x26x+92=0
(iv)3x24x+1=0
(v)(x+4)28x=0
(vi)(x2)22(x+1)=0
(vii)2x232x+12=0
(viii)x(1x)2=0
(ix)(x1)(x+2)+2=0
(x)(x+1)(x2)+x=0

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Answer:

(i) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form ax2+bx+c=0 Where a, b and c are real numbers with a0
The given equation is x23x+4=0
Here a=1,b=-3,c=4
D=b24ac=(3)24(1)(4)=916=7b24ac<0
Hence the equation has no real roots.
(ii) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form ax2+bx+c=0 Where a, b and c are real numbers with a0
The given equation is 2x2+x1=0
Here a=2,b=-1,c=-1
D=b24ac=(1)24(2)(1)=1+8=9b24ac>0
Here for the equation 2x2+x1=0, b24ac>0
Hence it is a quadratic equation with two distinct real roots
(iii) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form ax2+bx+c=0 Where a, b and c are real numbers with a0
The given equation is 2x26x+92=0
Here a=2,b=6,c=92
D=b24ac=(6)24(2)(94)=3636=0b24ac=0
here b24ac=0
Hence the equation have two real and equal roots.
(iv) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form ax2+bx+c=0 Where a, b and c are real numbers with a0
The given equation is 3x24x+1=0
Here a=3,b=-4,c=1
D=b24ac=(4)24(3)(1)=1612=4b24ac>0
Here b24ac>0
Hence the equation has distinct and real roots.
(v) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form ax2+bx+c=0 Where a, b and c are real numbers with a0
The given equation is (x+4)28x=0
(x)2+(4)2+2(x)(4)8x=0 (using(a+b)2=(a)2+(b)2+2ab)
(x)2+16+8x8x=0x2+16=0
a=1,b=0,c=16=(0)24(1)(16)=64b24ac<0
Here b24ac<0
Hence the equation has no real roots.
(vi) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form ax2+bx+c=0 Where a, b and c are real numbers with a0
The given equation is (x2)22(x+1)=0
(x)2+(2)2+2(x)(2)2x2=0 (using(a+b)2=(a)2+(b)2+2ab)
(x)2+222x2x2=0x2(22+2)x=0
Compare with ax2+bx+c=0 where a0
a=1,b=(22+2),c=0b24ac=((22+2))24(1)(0)=(22+2)20
=(22)2+(2)2+2(22)(2) (using(a+b)2=(a)2+(b)2+2ab)
=8+4+82=12+82
b24ac>0
Here b24ac>0
Hence the equation has two distinct real roots.
(vii) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form ax2+bx+c=0 Where a, b and c are real numbers with a0
The given equation is 2x232x+12=0
Here a=2,b=32,c=12
b24ac=(32)24(2)(12)=9222=4.53.8=1.7b24ac>0
Here b24ac>0
Hence the equation has distinct real roots.
(viii) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form ax2+bx+c=0 Where a, b and c are real numbers with a0
The given equation is x(1x)2=0
x(x)22=0x2+x2=0
compare with ax2+bx+c=0 where a0
a=1,b=1,c=2b24ac=(1)24(1)(2)18=7b24ac<0
Here b24ac<0
Hence the equation has no real roots.
(ix) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form ax2+bx+c=0 Where a, b and c are real numbers with a0
The given equation is (x1)(x+2)+2=0
x(x+2)1(x+2)+2=0x2+2xx2+2=0x2+x=0
a=1,b=1,c=0b24ac=(1)24(1)(0)=1
b24ac>0
Here b24ac>0
Hence the equation has two distinct real roots.
(x) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form ax2+bx+c=0 Where a, b and c are real numbers with a0
The given equation is (x+1)(x2)+x=0
x(x2)+1(x2)+x=0x22x+x2+x=0x22x+2x2=0x22=0
Compare with ax2+bx+c=0 where a0
a=1,b=0,c=2b24ac=(0)24(1)(2)=8
b24ac>0
Here b24ac>0
Hence the equation has two distinct real roots.

Question:2

Write whether the following statements are true or false. Justify your answers.
(i) Every quadratic equation has exactly one root.
(ii) Every quadratic equation has at least one real root.
(iii) Every quadratic equation has at least two roots.
(iv) Every quadratic equations has at most two roots.
(v) If the coefficient of x2 and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.
(vi)If the coefficient of x2 and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots.

Answer:

(i) False
Solution
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, ax2+bx+c=0
Where a, b and c are real numbers with a0
Root: If ax2+bx+c=0 …..(1)
Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.
Let us consider a quadratic equation x24=0
x24=0x2=4x=±4x=2x=2
Here – 2, 2 are the two roots of the equation.
Hence it is false that every quadratic equation has exactly one root.
(ii) False
Solution
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, ax2+bx+c=0
Where a, b and c are real numbers with a0
Root: If ax2+bx+c=0 …..(1)
Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.
Let us consider a quadratic equation
x2x+2=0
x2x+2=0
compare with ax2+bx+c=0 where a0
a=1,b=1,c=2b24ac=(1)24(1)(2)=18=7b24ac<0
Hence both the roots of the equation are imaginary.
Hence the statement every quadratic equation has at least one real root is False.
(iii) False
Solution
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, ax2+bx+c=0
Where a, b and c are real numbers with a0
Root: If ax2+bx+c=0 …..(1)
Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.
Let us consider a quadratic equation x24x+4=0
x24x+4=0x2+(2)22(x)(2)(x2)2=0(using(ab)2=a2+b22ab)(x2)=0x=2
The equation x24x+4=0 has only one root which is x = 2
Hence the given statement is False
(iv) True
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, ax2+bx+c=0
Where a, b and c are real numbers with a0
Root: If ax2+bx+c=0 …..(1)
Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.
As we know that the standard form of a quadratic equation is ax2+bx+c=0
It is a polynomial of degree 2.
As per the power of x is 2. There is at most 2 values of x exist that satisfy the equation.
Hence the given statement every quadratic equation has at most two roots is true.
(v) True
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, ax2+bx+c=0
Where a, b and c are real numbers with a0
Root: If ax2+bx+c=0 …..(1)
Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.
Let us suppose a quadratic equation x2+x2=0
x2+x2=0 is a quadratic equation in which the coefficient of x2 and constant term have opposite signs.
x2+x2=0
compare with ax2+bx+c=0 where a0
a=1,b=1,c=2b24ac=(1)24(1)(2)=1+8=9b24ac>0
Here b24ac>0
Hence these types of equations have real roots.
(vi) True
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, ax2+bx+c=0
Where a, b and c are real numbers with a0
Root: If ax2+bx+c=0 …..(1)
Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.
Let us suppose a quadratic equation x2+4=0
In x2+4=0, the coefficient of x2 and the constant term has the same sign and coefficient of x is 0.
x2+4=0
Compare with ax2+bx+c=0 where a0
a=1,b=0,c=4b24ac=(0)24(1)(4)=16b24ac<0
Here b24ac<0
Hence these types of equations have no real roots.

Question:3

A quadratic equation with integral coefficient has integral roots. Justify your answer

Answer:

False
Solution
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, ax2+bx+c=0
Where a, b and c are real numbers with a0
Root: If ax2+bx+c=0 …..(1)
Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.
Let us take a quadratic equation with integral coefficient
2x23x5=0
compare with ax2+bx+c=0 where a0
(a=2,b=-3,c=-5)
let us find the roots of the equation
x=b+b24ac2ax=3±94(2)(5)2×2x=3±9+404x=3+494x=3+74=57x=374=1
Here we found that 5/2 is not on integral
Hence the given statement is false

Question:4

Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.

Answer:

True
Let us suppose a quadratic equation with rational coefficients.
x22x+4=0
compare with ax2+bx+c=0 where a0
a=-1,b=-2,c=4
let us find the roots of the equation
x=b+b24ac2ax=2±44(1)(4)2(1)x=2±202x=2±252x=2+252x=2252
Both of its roots are irrational
Hence the given statement is true

Question:5

Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rational? Why?

Answer:

True
Let us consider an equation with distinct irrational numbers
2x252x+42=0
compare with ax2+bx+c=0 where a0
a=2,b=52,c=42
let us find the roots of the equation
x=b+b24ac2ax=52±504(2)(42)22x=52±503222x=52±1822x=2(5±3)22=5±32x=5+32=4x=532=1
The roots are rational. Hence the given statement is true

Question:6

Is 0.2 a root of the equation x2 – 0.4 = 0? Justify

Answer:

False
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, ax2 + bx + c = 0 Where a, b and c are real numbers with a 0
Roots : If ax2 + bx + c = 0 …..(1)
Is a quadratic equation then the values of x which satisfy equation 1 are the roots of the equation.
Here the given equation is x2 – 0.4 = 0 …..(2)
If 0.2 is a root of equation 2 then it should satisfy its equation
Put x = 0.2 in (2)
(0.2)20.4=0(210)0.4=041000.4=00.040.4=00.360
Here 0.2 is not satisfying the equation (2)
Hence 0.2 is not a root of the equation x20.4=0

Question:7

If b = 0, c < 0, is it true that the roots of x2 + bx+ c = 0 are numerically equal and opposite in sign? Justify.

Answer:

True
Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, ax2 + bx + c = 0 Where a, b and c are real numbers with a 0
Roots : If ax2 + bx + c = 0 …..(1)
Is a quadratic equation then the values of x which satisfy equation 1 are the roots of the equation.
Here the given equation is x2 + bx+ c = 0 …..(2)
It is also given that b = 0, c < 0.
Let c=-y
Put b = 0, c = – y in (2)
x2+0(x)y=0x2=yx=±yx=+yx=y
Hence both the roots are equal and opposite in sign. Hence the given statement is true.

Question:1

Find the roots of the quadratic equations by using the quadratic formula in each of the following:
(i)2x23x5=0
(ii)5x2+13x+8=0
(iii)3x2+5x+12=0
(iv)x2+7x10=0
(v)x2+22x6=0
(vi)x2+35x+10=0
(vii)12x211x+1=0

Answer:

(i) 52,1
2x23x5=0
Compare with ax2+bx+c=0 where a0
Here a=2, b=-3, c=-5
x=b±b24ac2ax=(3)±(3)24(2)(5)2×2x=3±9+404x=3±74x=3+74=104=52x=374=44=1
x=52,1
(52,1) are the roots of the equation 2x23x5=0

(ii) 1,85
5x2+13x+8=0
Compare with ax2+bx+c=0 where a0
Here a=5, b=13, c=8
x=b±b24ac2ax=13±(13)24(5)(8)2×5x=13±16916010x=13±310x=13+310=1010=1x=13310=1610=85
x=1,85
(1,85) are the roots of the equation 5x2+13x+8=0
(iii) 43,3
3x2+5x+12=0
Compare with ax2+bx+c=0 where a0
Here a=-3, b=5, c=12
x=b±b24ac2ax=5±(5)24(3)(12)2×(3)x=5±25+1446x=5±136x=5+136=86=43x=5136=186=3
x=43,3
(43,3) are the roots of the equation 3x2+5x+12=0
(iv) 5,2
x2+7x10=0
Compare with ax2+bx+c=0 where a0
Here a=-1, b=7, c=-10
x=b±b24ac2ax=7±494(1)(10)2×(1)x=7±92x=7±32x=7+32=42=2x=732=102=5
x=5,2
(5,2) are the roots of the equation x2+7x10=0
(v) 2,32
x2+22x6=0
Compare with ax2+bx+c=0 where a0
Here a=1,b=22,c=6
x=b±b24ac2ax=22±(22)24(1)(6)2×1x=22±8+242x=22±422x=2(2±22)2x=2±22x=2+22=2x=222=32
x=2,32
2,32 are the roots of the equation x2+22x6=0
(vi) 25,5
x2+35x+10=0
Compare with ax2+bx+c=0 where a0
a=1,b=35,c=10
x=b±b24ac2ax=35±(35)24(1)(10)2×1x=35±45402x=35±52x=35+52=452=25x=3552=252=5
x=25,5
25,5 are the roots of the equation x2+35x+10=0
(vii) 11+3,113
12x211x+1=0
Compare with ax2+bx+c=0 where a0
Here a=12,b=11,c=1
x=b±b24ac2ax=11±(11)24(12)(1)2×12x=11±1121x=11±3x=11+3x=113
x=(11+3),(113)
hence (11+3),(113) are the root of the equation 12x211x+1=0

Question:2

Find the roots of the following quadratic equations by the factorisation method
(i)2x2+53x2=0
(ii)25x2x35=0
(iii)32x25x2=0
(iv)3x2+55x10=0
(v)21x22x+121=0

Answer:

(ii) 3,12
Solution
25x2x35=02x25x35=02x25x3=02x26x+x3=02x(x3)+1(x3)=0(x3)(2x+1)=0x3=02x+1=0x=32x=1x=12
Hence 32,23 are the roots of the equation.
(i) 32,23
Solution
2x2+53x2=06x2+5x63=06x2+5x6=06x2+9x4x6=03x(2x+3)2(2x+3)=0(2x+3)(3x2)=02x+3=03x2=02x=33x=2x=32x=23
Hence 3,12 are the roots of the equation.
(iii) 2,132
Solution
32x25x2=032x26x+x2=032x(x2)+1(x2)=0(x2)(32x+1)=032x+1=0x2=032x=1x=2x=132
Hence 2,132 are root of the equation 32x25x2=0
(iv) 25,53
Solution
3x2+55x10=03x2+65x5x10=03x(x+25)5(x+25)=0(3x5)(x+25)=03x5=5x+25=03x=5x=25x=53
Hence 25,53 are root of the equation 3x2+55x10=0
(v) 121
solution
21x22x+121=021x2xx+121=0x(21x1)121(21x1)=0(21x1)(x121)=021x=0x121=021x=1x=121x=121
Hence 121 are the roots of the equation 21x22x+121=0

Question:8

At t minutes past 2 pm, the time needed by the minutes hand of a clock to show 3 pm was found to be 3 minutes less than t24 minutes. Find t.

Answer: 14

At t minutes past 2 pm, the time need by minute hand to show 3 pm = 60 – t
According to question, it is given that (60 – t) is equal to 3 minutes less than t24
60t=t24360t=t2124t212=4(60t)t212=2404tt2+4t12240=0t2+4t252=0t2+18t14t252=0t(t+18)14(t+18)=0(t+18)(t14)=0t=18,14
Time can’t equal to negative term.
Hence t = 14 minutes

Question:7

In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m2 [see Fig. 4.1]. Find the length and breadth of the pond.
447

Answer:

length = 34m, breadth = 24m
Solution
Let the distance between the pond and loan boundary is x.
Hence the length of pond =50-x-x
Then breadth of pond =40-x-x
From the figure
Area of lawn = Area of pond + Area of grass
50 × 40=(50-2x)(40-2x)+1184 (using Area of rectangle = l × B)
2000-1184=2000-100x-80x+4x2
4x2-180x+2000-2000+1184=0
4x2-180x+1184=0
Divide by 4
x2-45x+296=0
x2 - 37x - 8x + 296 = 0
x(x-37)-8(x-37)=0
(x-37)(x-8)=0
x = 37, 8
If x = 37
Length of pond =50-2(37)
=50-74
=-24
Which is not possible because length can’t be negative.
If x = 8
Length of pond =50-2(8)
=50-16=34m
Breadth of pond =40-2(8)
=40-16=24m

Question:6

At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.

Answer:

5, 27
Solution
Let Nisha’s present age = x
At present Asha is 2 more than the square of Nisha’s age.
Hence, Asha present age = x2 + 2
According to question
x2 + 2 - x = 10x – 1 - ( x2 + 2 )
x2 + 2 – x = 10x – 1 – x2 - 2
x2 + 2 – x - 10x + x2 + 3=0
2x2-11x+5=0
2x(x-5)-1(x-5)=0
(x-5)(2x-1)=0
x=5,12?
If Nisha’s age is 5
Asha age = (5)2+2=27
If Nisha’s age is 1/2 then Asha’s age is 2.25 which is not possible.

Question:5

If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?

Answer: 14

Let zeba’s current age = x
According to question
(x5)2=11+5xx2+2510=11+5x\(using(ab)2=a2+b22ab)x2+2510x115x=0x215x+14=0x214xx+14=0x(x14)1(x14)=0(x14)(x1)
x = 14
we can’t take 1 because if we take 5 years younger of 1 then it is equal to – 4 and age cannot be negative.
Hence x = 14 i.e., the age of zeba is 14 years

Question:4

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find the original speed of the train.

Answer:

45 km per hour.
Solution
Let the original speed of the train = x km per hour.
Total distance = 360 km
Time taken by train to reach at final point =(360x)(speed=distancetime)
If train’s speed increased by 5 then speed = (x + 5) km per hour. Time taken by train to cover 360 km at a speed of (x + 5) km per hour=
(360x+5)(speed=distancetime)
Now it is given that the difference between the time taken at a speed of x and at a speed of x + 5 is 48 min.
360x360x+5=4860(hour)360(x+5)360xx(x+5)=4860(hour)5(360x+1800360x)=4x(x+5)9000=4x2+20x4x2+20x9000=0
Or
x2+5x2250=0 (divide by 4)
x2+5x2250=0x2+50x45x2250=0x(x+50)45(x+50)=0x=50,45
We can’t take x = – 50 because speed is always positive.
Hence original speed of train is 45 km per hour.

Question:3

A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.

Answer:8

Let the natural number = x
It is given that x is increased by 12, equal to 160 times its reciprocal
x+12=160xx(x+12)=160x2+12x160=0x2+20x8x160=0x(x+20)8(x+20)=0(x+20)(x8)=0x+20=0x8=0x=20x=8
We can’t take x = – 20 because it is given that x is a natural number.
Hence x = 8 is the required natural number.

Question:2

Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number

Answer:

12
Let the natural number = x
It is given that when the square of x is diminished by 84 then it is equal to thrice of 8 more than x
x284=3(x+8)x284=3x+24x23x8424=0x23x108=0x212x+9x108=0x(x12)+9(x12)=0(x12)(x+9)=0
x = 12, – 9
We can’t take – 9 because it is given that x is a natural number.
Hence x = 12 is the required natural number.

Question:1

Find whether the following equations have real roots. If real roots exist, find them.
(i)8x2+2x3=0
(ii)2x2+3x+2=0
(iii)5x22x10=0
(iv)12x3+1x5=1,x32,5
(v)x2+55x70=0

Answer:

(i) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0 in the case of imaginary roots.
For the equation 8x2+2x3=0
Compare with ax2+bx+c=0 where a0
a=8,b=2,c=-3
b24ac=(2)24(8)(3)=4+96=100b24ac>0
Hence the equation has real roots
8x2+2x3=08x2+6x4x3=02x(4x+3)1(4x+3)=0(4x+3)(2x1)=04x+3=02x1=04x=32x=1x=34x=12
Roots of the equation 8x2+2x3=0 are 34,12
(ii) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0 in the case of imaginary roots.
For the equation 2x2+3x+2=0
Compare with ax2+bx+c=0 where a0
a=-2,b=3,c=2
b24ac=(3)24(2)(2)=9+16=25b24ac>0
Hence the equation has real roots
2x2+3x+2=02x2+4xx+2=02x(x2)1(x2)=0(x2)(2x1)=02x1=0x2=02x=1x=2x=12
Roots of the equation 2x2+3x+2=0 are 2,12
(iii) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0 in the case of imaginary roots.
For the equation 5x22x10=0
Compare with ax2+bx+c=0 where a0
a=5,b=-2,c=-10
b24ac=(2)24(5)(10)=4+200=204b24ac>0
Hence the equation has real roots
5x22x10=0x=b±b24ac2ax=2±44(5)(10)2(5)x=2±4+20010x=2±20410x=2(1±51)10x=1±515x=1+515,x=1515
Roots of the equation 5x22x10=0 are 1+515,1515
(iv) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0 in the case of imaginary roots.
For the equation 12x3+1x5=1
(x5)+(2x3)(2x3)(x5)=1x5+2x3=2x(x5)3(x5)3x8=2x210x3x+152x213x+153x+8=02x216x+23=0
Compare with ax2+bx+c=0 where a0
a=2,b=-16,c=23
b24ac=(16)24(2)(23)=256184=72b24ac>0
Hence the equation has real roots
x=b±b24ac2ax=16±722(2)(b24ac=72)x=16±624=8±322x=8+322,8322
Roots of the equation 12x3+1x5=1 are 8+322,8322
(v) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0 in the case of imaginary roots.
For the equation x2+55x70=0
Compare with ax2+bx+c=0 where a0
a=1,b=55,c=70b24ac=(55)24(1)(70)=125+280=405b24ac>0
Hence the equation has real roots
x2+55x70=0x2+75x25x70=0x(x+75)25(x+75)=0(x+75)(x25)x+75=0x+75=0x=75x=25
Roots of the equation x2+55x70=0 are 75,25

Question:1

Choose the correct answer from the given four options in the following questions: Which of the following is a quadratic equation?
(A)x2+2x+1=(4x)2+3(B)2x2=(5x)(2x25)(C)(k+1)x2+32x=7wherek=1(D)x3x2=(x1)3

Answer:

Answer(D)x3x2=(x1)3
Solution:
Quadratic equation:
A quadratic equation in x is an equation that can be written in the standard form.
ax2+bx+c=0 where a, b and c are real number with a0
(A)x2+2x+1=(4x)2+3x2+2x+1=(4)2+(x)22(4)(x)+3(Using(ab)2=a2+b22ab)x2+2x+1=16+x28x+3x2+2x+116x2+8x3=010x18=0
10x18=0 is not in the form of ax2+bx+c=0
Hence it is not a quadratic equation.
(B)
2x2=(5x)(2x25)2x2=10x22x2+25x10x+25x2=050x+2x105=05x10=0
This is not in the form of ax2+bx+c=0
Hence it is not a quadratic equation
(C)(k+1)x2+32x=7wherek=1
Put k = – 1
((1)+1)x2+32x=732x7=03x142=03x14=0
This is not in the form of ax2+bx+c=0
Hence this is not a quadratic equation.
(D)
x3x2=(x1)3x3x2=(x)3(1)3(1)+3(x)(1)2[using(ab)3=a3b33a2b+3ab2]x3x2x3+1+3x23x=02x23x+1=0
This is in the form of ax2+bx+c=0
Hence x3x2=(x1)3 is a quadratic equation.

Question:2

Which of the following is not a quadratic equation?
(A)2(x1)2=4x22x+1(B)2xx2=x2+5(C)(2x+3)2+x2=3x25x(D)(x2+2x)2=x4+3+4x3

Answer:

(C)(2x+3)2+x2=3x25x
Quadratic equation: A quadratic equation in x is an equation that can be written in the standard form ax2+bx+c=0, a0
Where a, b and c are real numbers
2(x1)2=4x22x+1
2(x2+12x)=4x22x+1
[using (a – b)2 = a2 + b2 – 2ab]
2x2+24x4x2+2x1=02x22x+1=0
It is a quadratic equation because it is in the form
ax2+bx+c=0
(B)
2xx2=x2+52xx2x25=02x2+2x5=02x2+2x5=0
It is a quadratic equation because it is in the form ax2+bx+c=0
(C)
(2x+3)3+x2=3x25x(2x)2+(3)2+2(2x)(3)+x2=x25x [using(a+b)2=a2+b2+2ab]
2x2+3+26x+x23x2+5x=0(26+5)x+3=0
It is not in the form of ax2+bx+c=0
Hence it is not a quadratic equation.
(D)
(x2+2x)2=x4+3+4x3(x2)2+(2x)2+2(x2)(2x)=x4+3+4x3 [using(a+b)2=a2+b2+2ab]
x4+4x2+4x3x434x3=04x23=0
It is in the form of ax2+bx+c=0
Hence it is a quadratic equation
Only option (C) is not in the form of ax2+bx+c=0
Hence (C) is not a quadratic equation.

Question:3

Which of the following equations has 2 as a root ?
(A)x24x+5=0(B)x2+3x12=0(C)2x27x+6=0(D)3x26x2=0

Answer:

(C)(C)2x27x+6=0
Solution
(A) x24x+5=0
Put x = 2
(2)24(2)+5=048+5=098=010
(B)
x2+3x12=0putx=2(2)2+3(2)12=04+612=01012=020
(C)
2x27x+6=0putx=22(2)27(2)+6=0814+6=01414=00=0
(D)
3x26x2=0putx=23(2)26(2)2=012122=020
Hence option (C) is only satisfy x = 2.
Hence C has 2 as a root.

Question:4

If 12 is a root of the equation x2+kx54=0 , then the value of k is
(A)2(B)2(C)14(D)12

Answer: (A) 2

As we know that if 12 is a root of the equation x2+kx54=0 then x=12 should satisfy its equation.
Put x=12
(12)2+k(12)54=014+k254=01+2k54=02k4=02k=4k=42=2k=2
Hence value of k is 2.

Question:5

Which of the following equations has the sum of its roots as 3?
(A)2x23x+6=0(B)x2+3x3=0(C)2x232x+1=0(D)3x23x+3=0

Answer:

(B) (B)x2+3x3=0
If (B)ax2+bx+c=0 is a quadratic equation the sum of its roots are ba
(A)
2x23x+6=0
Here b=-3,a=2
Sum of its roots =ba = (32)=32
(B)
x2+3x3=0
Here b=-3,a=3
Sum of its roots =ba=31=3
(C)
2x232x+1=0
Here b=32,a=2
Sum of its roots = ba=(32)×2=32
(D)
3x23x+3=0
Here b=-3,a=3
Sum of its roots =ba=(3)3=33=1
Hence only B has sum of its roots 3

Question:6

Values of k for which the quadratic equation 2x2kx+k=0 has equal roots is
(A) 0 only (B) 4 (C) 8 only (D) 0, 8

Answer:

(D) 0, 8
Here the given equation is 2x2kx+k=0
Compare with ax2+bx+c=0 where a0
a=2,b=-k,c=k
It is given that roots are equal.
Hence
b24ac=0(k)24(2)(k)=0k28k=0k(k8)=0k8=0k=0k=8k=0,8
Hence for the value of k = 0, 8 the equation ax2+bx+c=0 has equal roots.

Question:7

Which constant must be added and subtracted to solve the quadratic equation9x2+34x2=0 by the method of completing the square?
(A)18(B)164(C)14(D)964

Answer:

Here the given quadratic equation is
9x2+34x2=0
(3)2+14(3x)2=0(3x)2+2×(18)×(3x)2=0
[add and subtract(18)2]
(3x)2+2×(18)×(3x)+(18)2(18)22=0(3x)2+2×(18)×(3x)+(18)2=(18)2+2a2+b2+2ab=(a+b)2
So
(3x+18)2=1+64×264  Hence(18)2=164  should be added and subtracted in the equation 9x2+34x2=0
 for solving by completing the square method. 

Question:

The quadratic equation 2x25x+1=0 has
(A) two distinct real roots (B) two equal real roots
(C) no real roots (D) more than 2 real roots

Answer:

(C) no real roots
Here the given quadratic equation is 2x25x+1=0
 Compare with ax2+bx+c=0  where  a0
a=2,b=5,c=1b24ac=(5)24(2)(1=58=3
b24ac<0
Hence the equation 2x25x+1=0 has no real roots.

Question:9

Which of the following equations has two distinct real roots?
(A)2x232x+94=0(B)x2+x5=0(C)x2+3x+22=0(D)5x23x+1=0

Answer:

We know that if roots are distinct and real, then b24ac>0
(A)2x232x+94=0
compare with  ax2+bx+c=0 where a0
a=2,b=32,c=94
b24ac=(32)24(2)(94)=9×218=1818=0b24ac=0(two equal real roots)
(B) x2+x5=0
b24ac=(1)24(1)(5)=1+20=21b24ac>0(two distinct real roots) 
(C) x2+3x+22=0

\\b^{2}-4ac=(3 )^2-4(1) ( 2\sqrt{2} )\\ =9 -8\sqrt{2}\\ =9-11.3=-2.3\\ b^{2}-4ac<0\text{(no real roots)}(D) 5x23x+1=0
b24ac=(3)24(5)(1)=920=11b24ac<0(no real roots)
Here only option (B) have two distinct real roots.

Question:11

(x2+1)2x2=0 has
(A) four real roots (B) two real roots
(C) no real roots (D) one real root.

Answer:

(C) no real roots
Solution
Here the given equation is (x2+1)2x2=0
(x2)2+(1)2+2(x2)(1)x2=0 [using? ]
x4+1+2x2x2=0x4+x2+1=0
Put x2=z
z2+z+1=0
Compare with az2+bz+c=0 where a0
Here a=1,b=1,c=1
b24ac=(1)24(1)(1)=14=3b24ac<0
Here the given equation has no real roots.


NCERT Exemplar Solutions Class 10 Maths Chapter 4 Important Topics:

  • How to judge the nature of roots by observing the quadratic equation.
  • Students will learn different techniques to solve the quadratic equation such as factorisation or making perfect square etc.
  • NCERT exemplar Class 10 Maths solutions chapter 4 discusses the methods to form a quadratic equation as mentioned in the CBSE Class 10 Maths Syllabus. For example: sum of a number and its reciprocal is equal to 5. Find out the number.” This information will turn into a quadratic equation of variable that represents the number.
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NCERT Class 10 Exemplar Solutions for Other Subjects:

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As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 10 Maths Solutions Chapter 4:

These Class 10 Maths NCERT exemplar chapter 4 solutions provide a detailed knowledge of quadratic equations. The knowledge of quadratic equations will be useful even for students who do not pursue mathematics in higher classes. For students aspiring for engineering entrance examinations such as JEE Advanced, this chapter will be very important. Quadratic Equations practice problems of competitive level are also provided in the exemplar which can help the student to prepare well for all sorts of examinations.

The Class 10 Maths NCERT exemplar solutions chapter 4 Quadratic Equations cover a wide range of practice problems and solving them can help to excel in other books such as NCERT Class 10 Maths, RD Sharma Class 10 Maths, RS Aggarwal Class 10 Maths et cetera.

Students can easily download the pdf version of these solutions using NCERT exemplar Class 10 Maths solutions chapter 4 pdf download to review these detailed solutions while studying NCERT exemplar Class 10 Maths chapter 4.

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Also, Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. Does every quadratic equation have two roots?

No, it is not necessary for any quadratic equation to have two roots. Quadratic equation can have a unique root as well as some quadratic equations cannot have any roots or zeros.

2. What is the importance of discriminant in quadratic equation?

The discriminant of quadratic equations is very important to judge the nature of roots.

By evaluating the value of discriminant, we can depict that the given quadratic equation will have any real root or not.

3. Is the chapter Quadratic Equations important for Board examinations

The chapter of Quadratic Equations is important for Board examinations as it holds around 4-5% weightage of the whole paper.

4. Is Quadratic equation chapter of extreme importance from a board examination perspective?

Quadratic Equation is one of the most important topics of Algebra of Class 10 and holds around 8-10% marks of the final paper. If a student practices and refers NCERT exemplar Class 10 Maths solutions chapter 4 he/she can score well in this vertical.  

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

https://medicine.careers360.com/neet-college-predictor

Hope this helps you .

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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