Chapter 4: Quadratic Equations

Ans: Given: Quadratic equations

To find: nature of the roots. 

We know that for a quadratic equation $a{x^2} + bx + c = 0$

Discriminant $ = {{\text{b}}^2} - 4{\text{ac}}$

(A) If ${b^2} - 4ac > 0 \to $ two distinct real roots

(B) If ${b^2} - 4ac = 0 \to $ two equal real roots

(C) If ${b^2} - 4ac < 0 \to $ no real roots

(I) $\quad 2{{\text{x}}^2} - 3{\text{x}} + 5 = 0$

Comparing the given equation with ${\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}} = 0$, we obtain ${\text{a}} = 2,\;{\text{b}} =  - 3,{\text{c}} = 5$

Discriminant : ${b^2} - 4ac$ $ = {( - 3)^2} - 4(2)(5) = 9 - 40$

$ =  - 31$

As ${b^2} - 4ac < 0$

Therefore, no real root is possible for the given equation.

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

Comparing the given equation with $a{x^2} + bx + c = 0$, we obtain $a = 3,b =  - 4\sqrt 3 ,c = 4$

Discriminant $:{b^2} - 4ac = {( - 4\sqrt 3 )^2} - 4(3)(4)$

$ = 48 - 48 = 0$

As ${b^2} - 4ac = 0$

Therefore, real roots exist for the given equation and they are equal to each other. And the roots will be $\dfrac{{ - b}}{{2a}}$ and $\dfrac{{ - b}}{{2a}}$.

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

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

(III) $2{{\text{x}}^2} - 6{\text{x}} + 3 = 0$

Comparing the given equation with $a{x^2} + bx + c = 0$, we obtain ${\text{a}} = 2,\;{\text{b}} =  - 6,{\text{c}} = 3$

Discriminant $ = {b^2} - 4ac = {( - 6)^2} - 4(2)(3)$

$ = 36 - 24 = 12$

As ${b^2} - 4ac > 0$

Therefore, distinct real roots exist for the given equation as follows. $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

$ = \dfrac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4(2)(3)} }}{{2(2)}}$

$ = \dfrac{{6 \pm \sqrt {12} }}{4} = \dfrac{{6 \pm 2\sqrt 3 }}{4}$

$ = \dfrac{{3 \pm \sqrt 3 }}{2}$

Therefore, the root are $\dfrac{{3 + \sqrt 3 }}{2}$ or $\dfrac{{3 - \sqrt 3 }}{2}$

Ans: Given: A quadratic equation

To find: k if the roots are equal 

We know that if an equation $a{x^2} + bx + c = 0$ has two equal roots, its Discriminant $\left( {{b^2} - 4ac} \right)$ will be 0 .

(I) $2{{\text{x}}^2} + {\text{kx}} + 3 = 0$

Comparing equation with $a{x^2} + bx + c = 0$, we obtain

${\text{a}} = 2,\;{\text{b}} = {\text{k}},{\text{c}} = 3$

Discriminant $ = {b^2} - 4ac = {(k)^2} - 4(2)(3)$

$ = {{\text{k}}^2} - 24$

For equal roots, Discriminant $ = 0$

${k^2} - 24 = 0$

${k^2} = 24$

$k =  \pm \sqrt {24}  =  \pm 2\sqrt 6 $

(II) ${\text{kx}}({\text{x}} - 2) + 6 = 0$

or $k{x^2} - 2kx + 6 = 0$

Comparing the given equation with ${\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}} = 0$, we obtain $a = k,b =  - 2k,c = 6$

Discriminant $ = {b^2} - 4ac = {( - 2k)^2} - 4(k)(6)$

$ = 4{k^2} - 24k$

For equal roots, ${{\text{b}}^2} - 4{\text{ac}} = 0$

$4{{\text{k}}^2} - 24{\text{k}} = 0$

$4k(k - 6) = 0$

Either $4{\text{k}} = 0$ or ${\text{k}} = 6 = 0$

${\text{k}} = 0$ or ${\text{k}} = 6$

However, if ${\text{k}} = 0$, then the equation will not have the terms ' ${{\text{x}}^2}$ ' and ' ${\text{x}}$' .

Therefore, if the given equation has two equal roots, k should be 6 only.

Ans: Given: A rectangular mango grove.

To find: its dimensions

Let us consider the breadth of mango grove be $l$. 

Length of mango grove will be 2l . Area of mango grove $ = (2l)(l) = 2{l^2}$ $ = 800$

${l^2} = \dfrac{{800}}{2} = 400$

${l^2} - 400 = 0$

Comparing the given equation with ${\text{a}}{{\text{l}}^2} + {\text{bl}} + {\text{c}} = 0$, we obtain $a = 1 ,b = 0,c = 400$

Discriminant $ = {b^2} - 4ac = {(0)^2} - 4 \times (1) \times ( - 400) = 1600$

Here, ${b^2} - 4ac > 0$

Therefore, the equation will have real roots.

$l = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

$l = \dfrac{{ - 0 \pm \sqrt {1600}}}{{2*1}}$

And Therefore, the desired rectangular mango grove can be designed. $l =  \pm 20$

However, length cannot be negative. Therefore, breadth of mango grove $ = 20\;{\text{m}}$

Length of mango grove $ = 2 \times 20 = 40\;{\text{m}}$

Ans: Given: A situation.

To find: Predict the situation whether it is possible or not? 

Let us consider the age of one friend be $x$ years. 

Age of the other friend will be $(20 - {\text{x}})$ years.

4 years ago, age of 1 st friend $ = ({\text{x}} - 4)$ years 

And, age of ${2^{{\text{nd }}}}$ friend $ = (20 - {\text{x}} - 4)$ $ = (16 - {\text{x}})$ years

Given that, $(x - 4)(16 - x) = 48$

$16x - 64 - {x^2} + 4x = 48$

$ - {x^2} + 20x - 112 = 0$

${x^2} - 20x + 112 = 0$

Comparing the given equation with ${\text{a}}{{\text{x}}^2} + {\text{bx}} + {\text{c}} = 0$, we obtain ${\text{a}} = 1,\;{\text{b}} =  - 20,{\text{c}} = 112$

Discriminant $ = {b^2} - 4ac = {( - 20)^2} - 4(1)(112)$

$ = 400 - 448 =  - 48$

As ${b^2} - 4ac < 0$

Therefore, no real root is possible for the given equation and Therefore, the given situation is not possible.

Ans: Given: A situation.

To find: Predict the situation whether it is possible or not? 

Let us consider the length and breadth of the park be l and ${\text{b}}$. 

Perimeter $ = 2(l + b) = 80$ $l + b = 40$

Or, ${\text{b}} = 40 - l$

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

$40l - {l^2} = 400$

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

Comparing the given equation with ${\text{a}}{{\text{l}}^2} + {\text{bl}} + {\text{c}} = 0$, we obtain

${\text{a}} = 1,\;{\text{b}} =  - 40,{\text{c}} = 400$

Discriminate $ = {b^2} - 4ac = {( - 40)^2} - 4(1)(400)$

$ = 1600 - 1600 = 0$

As ${b^2} - 4ac = 0$

Therefore, the given equation has equal real roots. And Therefore, the given situation

is possible. Root of the given equation, $l =  - \dfrac{b}{{2a}}$

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

Therefore, length of park, $l = 20\;{\text{m}}$ 

And breadth of park, $b = 40 - l = 40 - 20 = 20\;{\text{m}}$