NCERT Solutions Class 10 Maths Chapter 4 for Exercise 4.3 - Quadratic Equations

Q1 (ii) Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: $3x^2 - 4\sqrt3x + 4 = 0$

Answer:

To check whether real roots exist, we first find the discriminant.

$D = b^2-4ac$

If D > 0, then the roots are distinct and real.

If D < 0, then no real roots.

If D = 0, then there exist two equal real roots.

Given the quadratic equation, $3x^2 - 4\sqrt3x + 4 = 0$

$b^2-4ac=(-4\sqrt{3})^2-(4\times4\times3)=48-48=0$

Here, the value of the discriminant = 0, which implies that roots exist and the roots are equal.

The roots are given by the formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\sqrt{3}\pm\sqrt{0}}{2\times3}=\frac{2}{\sqrt{3}}$

So the roots are

$\frac{2}{\sqrt{3}},\ \frac{2}{\sqrt{3}}$

Q1 (iii) Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: $2x^2 - 6x + 3 = 0$

Answer:

To check whether real roots exist, we first find the discriminant.

$D = b^2-4ac$

If D > 0, then the roots are distinct and real.

If D < 0, then no real roots.

If D = 0, then there exist two equal real roots.

Given the quadratic equation, $2x^2 - 6x + 3 = 0$

$b^2-4ac=(-6)^2-4\times2\times3=12$

The discriminant > 0. Therefore, the given quadratic equation has two distinct real roots

The roots are given by the formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-6\pm\sqrt{12}}{2\times2}=\frac{3}{2}\pm\frac{\sqrt{3}}{2}$

So the roots are

$\frac{3}{2}+\frac{\sqrt{3}}{2}, \frac{3}{2}-\frac{\sqrt{3}}{2}$

Q2 (i) Find the values of k for each of the following quadratic equations so that they have two equal roots; $2x^2 + kx + 3 = 0$

Answer:

For two equal roots for the quadratic equation: $ax^2+bx+c =0$, the value of the discriminant $ D=0$.

Given equation: $2x^2 + kx + 3 = 0$

Comparing and getting the values of a,b, and c.

$a = 2, \ b = k,\ c = 3$

The value of $D = b^2-4ac = (k)^2 - 4(2)(3)$

$\Rightarrow (k)^2 = 24$

Or, $\Rightarrow k=\pm \sqrt{24} = \pm 2\sqrt{6}$

Q2 (ii) Find the values of k for each of the following quadratic equations so that they have two equal roots $kx(x-2) + 6 = 0$

Answer:

For two equal roots for the quadratic equation: $ax^2+bx+c =0$, the value of the discriminant is $ D=0$.

Given equation: $kx(x-2) + 6 = 0$

Can be written as: $kx^2-2kx+6 = 0$

Comparing and getting the values of a,b, and c.

$a = k, \ b = -2k,\ c = 6$

The value of $D = b^2-4ac = (-2k)^2 - 4(k)(6) = 0$

$\Rightarrow 4k^2 - 24k = 0$

$\Rightarrow 4k(k-6) = 0$

$\Rightarrow k= 0\ or\ 6$

But $ k=0$ is NOT possible because it will not satisfy the given equation.

Hence, the only value of $k$ is 6 to get two equal roots.

Q3 Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m ²? If so, find its length and breadth.

Answer:

Let the breadth of the mango grove be $b$.

Then, according to the question, the length of the mango grove will be $ 2 b$.

Therefore, the area will be:

$Area = (2b)(b) = 2b^2$

Which will be equal to $ 800 m^2$ according to the question.

$\Rightarrow 2b^2 = 800m^2$

$\Rightarrow b^2 - 400 = 0$

Comparing to get the values of $a,b,c$.

$a=1, \ b= 0 , \ c = -400$

Finding the discriminant value:

$D = b^2-4ac$

$\Rightarrow 0^2-4(1)(-400) = 1600$

Here, $D>0$

Therefore, the equation will have real roots.

And hence finding the dimensions:

$\Rightarrow b^2 - 400 = 0$

$\Rightarrow b = \pm 20$

We know that a negative value is not possible; therefore, the value of the breadth ofthe mango grove will be 20m.

And the length of the mango grove will be: $= 2\times20 = 40 m$

Q4 Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Answer:

Let the age of one friend be $ x\ years$ and the age of another friend will be: $(20-x)\ years.$

Now, 4 years ago, their ages were $x-4\ years$ and $20-x-4 \ years$.

According to the question, the product of their ages in years was 48.

Therefore (x-4)(20-x-4) = 48

$\Rightarrow 16x-64-x^2+4x= 48$

$\Rightarrow -x^2+20x-112 = 0$ or $\Rightarrow x^2-20x+112 = 0$

Now, comparing to get the values of $a,\ b,\ c$.

$a = 1,\ b= -20,\ c =112$

Discriminant value $D = b^2-4ac = (-20)^2 -4(1)(112) = 400-448 = -48$

As $D < 0$ .

Therefore, there are no real roots possible for this given equation, and hence, this situation is not possible.

Q5 Is it possible to design a rectangular park of perimeter 80 m and area 400 m ²? If so, find its length and breadth.

Answer:

Let us assume the length and breadth of the park be $'l'\ and\ 'b'$ respectively.

Then, the perimeter will be $P = 2(l+b) = 80$

$\Rightarrow l+b = 40\ or\ b = 40 - l$

Now, the area of the park is:

$Area = l\times b = l(40-l) = 40l - l^2$

Given : Area = 400 m²

So, $40l - l^2 = 400$

$l^2 - 40l +400 = 0$

Comparing to get the values of a, b and c.

The value of the discriminant $D = b^2-4ac$

$\Rightarrow = b^2-4ac = (-40)^2 - 4(1)(400) = 1600 -1600 = 0$

As $D = 0$ .

Therefore, this equation will have two equal roots.

And hence the roots will be:

$l =\frac{-b}{2a}$

$l =\frac{-40}{2(1)} = \frac{40}{2} =20$

Therefore, the length of the park, $l =20\ m$ and breadth of the park $b = 40-l = 40 -20 = 20\ m$ .