Quadratic Equations - Solved Examples

Exercise 4.3

For a quadratic equation $ax^2 + bx + c = 0$

I. If $b^2 - 4ac > 0$, two distinct real roots exist
II. If $b^2 - 4ac = 0$, two equal real roots exist
III. If $b^2 - 4ac < 0$, no real roots exist


(i) $2x^2 - 3x + 5 = 0$

In this quadratic equation,
$a = 2, b = -3, c = 5$

$b^2 - 4ac$ = $(-3)^2 - 4 × 2 × 5$ = $9 - 40 = -31$

As $b^2 - 4ac < 0$, no real roots exist.



(ii) $3x^2 - 4\sqrt{3}x + 4 = 0$

In this quadratic equation,
$a = 3, b = -4\sqrt{3}, c = 4$

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

As $b^2 - 4ac = 0$, two equal real roots exist.


The roots will be $\dfrac{-b}{2a}$ and $\dfrac{-b}{2a}$

$\dfrac{-b}{2a}$ = $\dfrac{-(-4\sqrt{3})}{2 × 3}$ = $\dfrac{4\sqrt{3}}{6}$ = $\dfrac{2}{\sqrt{3}}$

Therefore, the roots are $\dfrac{2}{\sqrt{3}}$ and $\dfrac{2}{\sqrt{3}}$



(iii) $2x^2 - 6x + 3 = 0$

In this quadratic equation,
$a = 2, b = -6, c = 3$

$b^2 - 4ac$ = $(-6)^2 - 4 × 2 × 3$ = $36 - 24 = 12$

As $b^2 - 4ac > 0$, two distinct real roots exist.


The roots can be found out as

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ $ = \dfrac{-(-6) ± \sqrt{12}}{2 × 2}$ $ = \dfrac{6 ± 2\sqrt{3}}{4}$ $ = \dfrac{3 ± \sqrt{3}}{2}$

Therefore, the roots are $\dfrac{3 + \sqrt{3}}{2}$ and $\dfrac{3 - \sqrt{3}}{2}$

Find the value of $k$ for each of the following quadratic equations, so that they have two equal roots.

Solution:

A quadratic equation $ax^2 + bx + c = 0$ can have equal roots when
Discriminant $b^2 - 4ac = 0$


(i) $2x^2 + kx + 3 = 0$

Here $a = 2, b = k, c = 3$
∴ $b^2 - 4ac$ = $(k)^2 - 4 × 2 × 3$ = $ = k^2 - 24$

For roots to be equal, $b^2 - 4ac = 0$
∴ $k^2 - 24 = 0$
⇒ $k^2 = 24$
⇒ $k = ± \sqrt{24} = ± 2\sqrt{6}$



(ii) $kx(x - 2) + 6 = 0$

$kx(x - 2) + 6 = 0$
⇒ $kx^2 - 2kx + 6 = 0$

Here $a = k, b = -2k, c = 6$
∴ $b^2 - 4ac$ = $(-2k)^2 - 4 × k × 6 = 4k^2 - 24k$

For roots to be equal, $b^2 - 4ac = 0$
∴ $4k^2 - 24k = 0$
⇒ $4k(k - 6) = 0$
⇒ $k = 0$ or $k = 6$

If $k = 0$, the equation will no longer be quadratic.
Hence $k = 6$

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.

Solution:

Let the breadth of the rectangular mango grove be $x$ m
∴ it's length will be $2x$ m

Area = Length × Breadth
$800 = x × 2x$
⇒ $2x^2 = 800$
⇒ $x^2 = 400$
⇒ $x^2 - 400 = 0$

By comparing the above equation with quadratic equation $ax^2 + bx + c = 0$
$a = 1, b = 0, c = -400$
∴ $b^2 - 4ac$ = $(0)^2 - 4 × 1 × -400 = 1600$

As $b^2 - 4ac = 1600 > 0$, it is possible to design a mango grove.

$x^2 - 400 = 0$
$x^2 = 400$
$⇒ x = ± 20$

As sides of a rectangle cannot be negative, $x = 20$

Hence breadth is 20m and length = $2 × 20 = 40$m

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

Solution:

Let the age of first friend be $x$
∴ Age of second friend = $20 - x$

Four years ago,
Age of first friend = $x - 4$ and
Age of second friend = $20 - x -4$

The product of the ages, 4 years ago = $(x - 4)(20 - x - 4) = 48$
⇒ $(x - 4)(16 - x) = 48$
⇒ $16x - 64 - x^2 + 4x = 48$
⇒ $x^2 - 20x + 112 = 0$

Here $a = 1, b = -20, c = 112$
∴ $b^2 - 4ac$ = $(-20)^2 - 4 × 1 × 112 = 400 - 448 = -48$

As $b^2 - 4ac < 0$, no real roots exist.

So this situation is not possible.

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

Solution:

Perimeter of rectangular park = $2 (l + b) = 80$ -----(i)

Area of rectangular park = $l × b = 400$ -----(ii)

From equation (i),
$2 (l + b) = 80$
$⇒ 2l + 2b = 80$
$⇒ 2l = 80 - 2b$
$⇒ l = \dfrac{80 - 2b}{2}$
$⇒ l = 40 - b$

Substituting value of $l$ in equation (ii)
$(40 - b)b = 400$
⇒ $40b - b^2 = 400$
⇒ $b^2 - 40b + 400 = 0$

Comparing with equation $ax^2 + bx + c = 0$
$a = 1, b = -40, c = 400$
∴ $b^2 - 4ac$ = $(-40)^2 - 4 × 1 × 400 = 1600 - 1600 = 0$

∴ it is possible to design a rectangular park with the given conditions

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ $ = \dfrac{-(-40) ± \sqrt{0}}{2 × 1}$ $ = \dfrac{40}{2} = 20$

Length = $40 - b$ = $40 - 20 = 20$

Hence the rectangular park will be of length $= 20$ m and breadth $= 20$ m