Quadratic Equations ⇒⇒ Exercise 4.3

Question 1 (i)

Find the roots of the following quadratic equations, If the real roots exist, find them :
(i) $ 2x^2 – 3x + 5 = 0$

Solution :

The given equation is: $ 2x^2 – 3x + 5 = 0 $

Let, compare given quadratic equation with Standard Equation: ax² + bx + c = 0

In the given equation, a= 2 , b = -3 , and c = 5.

To check whether the roots of the given quadratic equation exist or not, we need to calculate the $ b^2 − 4ac $

$$ { (-3)^2 - (4 × 2 × 5) } $$

$$ ⇒{ 9 - 40} $$

$$ = -31 $$

Since, b2 - 4ac < 0.

Therefore, No real root is possible for the given equqtions $ 2x^2 – 3x + 5 = 0$

Question 1 (ii)

Find the roots of the following quadratic equations, If the real roots exist, find them :
(ii) $ 3x^2 – 4{\sqrt{3}}x + 4 = 0$

Solution :

The given equation is: $ 3x^2 – 4{\sqrt{3}}x + 4 = 0$

Let, compare given quadratic equation with Standard Equation: ax² + bx + c = 0

In the given equation, a= 3 , b = $- 4{\sqrt{3}} $, and c = 4 .

To check whether the roots of the given quadratic equation exist or not, we need to calculate the $ b^2 − 4ac $

$$ { -( 4{\sqrt{3}})^2 - (4 × 3 × 4) } $$

$$ ⇒ {(16 × 3) - 48} $$

$$ ⇒ 48 -48 $$

$$ = 0 $$

Since, b2 - 4ac = 0.

Therefore, There are two real and equal roots.

For equal roots, $$ x = - { b \over 2a} $$

$$ x = - { - 4{\sqrt {3} } \over {2 × 3}} $$

$$ x = { 2{\sqrt {3} } \over { 3}} $$

$$ x = { 2 \over { \sqrt {3}}} $$

Therefore, real root are = $ { 2 \over { \sqrt {3}}} , { 2 \over { \sqrt {3}}}$

Question 1 (iii)

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

Solution :

The given equation is: $ 2x ^2 - 6x + 3 = 0$

Let, compare given quadratic equation with Standard Equation: ax² + bx + c = 0

In the given equation, a = 2 , b = -6, and c = 3 .

To check whether the roots of the given quadratic equation exist or not, we need to calculate the $ b^2 − 4ac $

$$ { -6^2 - (4 × 2 × 3) } $$

$$ ⇒ 36 - 24 $$

$$⇒ 12 $$

Since, b2 - 4ac > 0.

Therefore, There are two real and distinct roots.

For distinct real roots,

By using quadratic formula , we obtain

$$ x = {{ - b \pm \sqrt{b^2 - 4ac}} \over {2a} }$$

Thus, roots of the given quadratic equation

$$ x = {{ -(- 6) \pm \sqrt{(-6^2) - 4 × 2 × 3}} \over {2×2} }$$

$$⇒ x = {{ 6 \pm \sqrt{(36) - 24}} \over {4} }$$

$$⇒ x = {{ 6 \pm \sqrt{12}} \over {4} }$$

$$⇒ x = {{ 6 \pm 2\sqrt{3}} \over {4} }$$

$$⇒ x = { 2({ 3 \pm \sqrt{3}}) \over {4} }$$

$$⇒ x = {{ 3 \pm \sqrt{3}} \over {2} }$$

Therefore, Roots of the equation are $ {{ 3 - \sqrt{3}} \over {2} }$ and $ {{ 3 + \sqrt{3}} \over {2} }$

Question 2 (i)

Find the value of k for the following quadratic equation, so that it has two equal roots.:
(i) $ 2x^2 + kx + 3 = 0$

Solution :

The given equation is: $ 2x^2 + kx + 3 = 0$

Let, compare given quadratic equation with Standard Equation: ax² + bx + c = 0

In the given equation, a= 2 , b = k , and c = 3 .

Since, the equation has two equal roots

D = 0

Since, b2 - 4ac = 0.

Therefore, There are two real and equal roots.

$$ {k^2 - (4×2×3)} = 0 $$

$$⇒ {k^2 - 24} = 0 $$

$$⇒ {k^2 = 24} $$

$$⇒ {k = \pm \sqrt{24} } $$

$$⇒ {k = \pm \sqrt{4×6} } $$

$$⇒ {k = \pm 2\sqrt{6} } $$

The required value of K are $ { 2\sqrt{6} }$ and $ {- 2\sqrt{6} }$

Question 2 (ii)

Find the value of k for the following quadratic equation, so that it has two equal roots.:
(ii) kx (x – 2) + 6 = 0

Solution :

The given equation is: kx (x – 2) + 6 = 0

$$ kx^2 - 2kx + 6 = 0 $$

Let, compare given quadratic equation with Standard Equation: ax² + bx + c = 0

In the given equation, a= k , b = -2k , and c = 6 .

Since, the equation has two equal roots

D = 0

Since, b2 - 4ac = 0.

Therefore, There are two real and equal roots.

$$ {(-2k)^2 - (4×k×6)} = 0 $$

$$⇒ {4k^2 - 24k} = 0 $$

$$⇒ {4k(k - 6) = 0 }$$

4k = 0 (or)⇒ k-6 = 0

k = 0 (or)⇒ k = 6

If we consider the value of k as 0, then the equation will not longer be quadratic.

Then required value of K =6

Question 3

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m² ? If so, find its length and breadth.

Solution :

Let, breadth of the rectangular grove = x meter

According to question, lenght = 2x meter

Given , Area of rectangular mango grove = 800 m²

Area = length × breadth

$$ 800 = x × 2x $$

$$ 800 = 2x^2 $$

$$ x^2 = 800 \over 2$$

$$ x^2 = 400 $$

$$ x = \pm{ \sqrt{400} } $$

$$ x = \pm{ 20 } $$

∴ x = 20 or x = −20

Value of x can’t be negative value as it represents the breadth of the rectangle.

∴ Possible Breadth of mango grove x = 20 meters

∴ Length of mango grove 2x = 2× 20 = 40 meters

Question 4

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.

Solution :

Let Age of first friend = x yrs

According to question,the sum of the ages of two friends are 20 years .

Present age of first friend + Present age of second friend = 20

'x' + Present age of second friend = 20

Present age of second friend = (20- x) yrs.

Four years ago

Age of first friend = (x - 4)yrs.

Age of second friend = (20 - x - 4) = (16 - x)yrs.

Thus, According to question:

Product of their ages 4 years ago was 48

$$ (x - 4) ×(16 - x) = 48 $$

$$⇒ x (16 - x) - 4(16 - x) = 48 $$

$$⇒ 16x - x^2 - 64 + 4x = 48 $$

$$⇒ - x^2 - 64 + 20x - 48 = 0 $$

$$⇒ - x^2 + 20x - 112 = 0 $$

$$⇒ x^2 - 20x + 112 = 0 $$

Let, compare given quadratic equation with Standard Equation: ax² + bx + c = 0

In the given equation, a = 1 , b = -20, and c = 112 .

To check whether the roots of the given quadratic equation exist or not, we need to calculate the D = $ b^2 − 4ac $

$$D = { (-20)^2 - (4 × 1 × 112) } $$

$$ ⇒ D = { 400 - 448} $$

$$ ⇒ D = -48 $$

Since, D = -48

Here, $ b^2 − 4ac $ < 0,

Thus ,It has no real roots of the given equation.

Question 5

Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth

Solution :

Let the length of rectangular park = x meter

Perimeter of the rectangular park = 80m

2 (length + breadth) = 80m

(length + breadth) = $ 80 \over 2$m

(x + breadth) = 40 m

breadth = (40- x )m

Given , Area of rectangular park = 400 m²

Area = length × breadth

$$ 400 = x × (40- x )$$

$$ 400 = 40x -x^2 $$

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

Let, compare given quadratic equation with Standard Equation: ax² + bx + c = 0

In the given equation, a= 1 , b = -40 , and c = 400.

To check whether the roots of the given quadratic equation exist or not, we need to calculate the D=b²−4ac

$$D = { (-40)^2 - (4 × 1 ×400) } $$

$$ ⇒D = { 1600 - 1600} = 0 $$

Since, D = 0

Hence, the given Quadratic equation has two equal real roots.

Now using quadratic formula to find roots

$$ x = {{ - b \pm \sqrt{D}} \over {2a} }$$

putting the values

$$ x = {{ - (-40) \pm \sqrt{0}} \over {2 × 1} }$$

$$ x = {40 \over {2} }$$

$$ x = {20} m $$

Thus ,length of rectangular park = 20 m .

And, Breadth of rectangular park = (40 – x)= (40 – 20) = 20m

Hence, condition is possible and length = 20 meters and breadth = 20 meters.