# CBSE 10th Mathematics | Nature of Roots of a Quadratic Equation

## Nature of Roots of a Quadratic Equation

In previous section, we have studied that the roots of the equation $\displaystyle a{{x}^{2}}+bx+c=0$ are given by

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

A quadratic equation $\displaystyle a{{x}^{2}}+bx+c=0$ has

·    Two distinct real roots if $\displaystyle {{b}^{2}}-4ac>0$.

If $\displaystyle {{b}^{2}}-4ac>0,$ we get two distinct real roots $\displaystyle -\frac{b}{2a}+\frac{\sqrt{{{b}^{2}}-4ac}}{2a}$ and $\displaystyle -\frac{b}{2a}-\frac{\sqrt{{{b}^{2}}-4ac}}{2a}$

·    Two equal roots, if $\displaystyle {{b}^{2}}-4ac=0$.

If $\displaystyle {{b}^{2}}-4ac=0,$ then $\displaystyle x=-\frac{b\pm 0}{2a}$

i.e.    $\displaystyle x=-\frac{b}{2a}$

So, the roots are both $\displaystyle -\frac{b}{2a}$

·    No real roots, if $\displaystyle {{b}^{2}}-4ac<0$

If $\displaystyle {{b}^{2}}-4ac<0,$ then there is no real number whose square is $\displaystyle {{b}^{2}}-4ac.$

Note: $\displaystyle \left( {{b}^{2}}-4ac \right)$ determines whether the quadratic equation $\displaystyle a{{x}^{2}}+bx+c=0$ has real roots or not, hence $\displaystyle \left( {{b}^{2}}-4ac \right)$ is called the discriminant of quadratic equation.

It is denoted by D.

## Solved Problems based on Nature of Roots of a Quadratic Equation

### Problem:

Find the discriminant of the quadratic equation $\displaystyle 2{{x}^{2}}-4x+3=0,$ and hence find the nature of its roots.

### Solution:

The given equation is of the form $\displaystyle a{{x}^{2}}+bx+c=0,$ where a = 2, b = – 4 and c = 3. Therefore, the discriminant.

\    $\displaystyle {{b}^{2}}-4ac={{(-4)}^{2}}-(4\times 2\times 3)=16-24=-8<0$

So, the given equation has no real roots.

### Problem:

Find the discriminant of the equation $\displaystyle 3{{x}^{2}}-2x+\frac{1}{3}=0$ and hence find the nature of its roots. Find them, if they are real.

### Solution:

Here $\displaystyle a=3,\,\,b=-2$ and $\displaystyle c=\frac{1}{3}.$

Therefore, discriminant $\displaystyle {{b}^{2}}-4ac={{(-2)}^{2}}-4\times 3\times \frac{1}{3}=4-4=0.$

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

The roots are $\displaystyle -\frac{b}{2a},-\frac{b}{2a},$ i.e., $\displaystyle \frac{2}{6},\frac{2}{6},$ i.e., $\displaystyle \frac{1}{3},\frac{1}{3}.$

### Problem:

Find the values of k for which the following equation has equal roots:

$\displaystyle (k-12)\,{{x}^{2}}+2(k-12)\,x+2=0$

### Solution:

We have,

$\displaystyle (k-12)\,{{x}^{2}}+2\,(k-12)\,x+2=0$

Here,    $\displaystyle a=k-12,\,\,b=2\,(k-12)$ and c = 2

\    $\displaystyle D={{b}^{2}}-4ac=4\,{{(k-12)}^{2}}-4\,\,(k-12)\times 2$

Þ    $\displaystyle D=4\,(k-12)\,\{(k-12)-2\}$

Þ    $\displaystyle D=4\,(k-12)\,(k-14)$

The given equation will have equal roots, if

$\displaystyle D=0$ Þ $\displaystyle 4\,(k-12)\,(k-14)=0$ Þ k – 12 = 0 or k – 14 = 0

Þ k = 12 or, k = 14

### Problem:

Find the values of k for which the given equation has real roots:

(i) $\displaystyle k{{x}^{2}}-6x-2=0$        (ii) $\displaystyle 9{{x}^{2}}+3kx+4=0$    (iii) $\displaystyle 5{{x}^{2}}-kx+1=0$

### Solution:

(i)    We have

$\displaystyle k{{x}^{2}}-6x-2=0$

Here,    $\displaystyle a=k,\,\,b=-6$ and c = –2

\    $\displaystyle D={{b}^{2}}-4ac={{(-6)}^{2}}-4\times k\times -2=36+8k$

The given equation will have real roots, if

$\displaystyle D\ge 0\Rightarrow 36+8k\ge 0\Rightarrow 8k\ge -36\Rightarrow k\ge \frac{-36}{8}\Rightarrow k\ge -\frac{9}{2}$

(ii)    The given equation is $\displaystyle 9{{x}^{2}}+3kx+4=0$

Here,    a = 9, b = 3k and c = 4

\    $\displaystyle D={{b}^{2}}-4ac=9{{k}^{2}}-4\times 9\times 4=9{{k}^{2}}-144$

The given equation will have real roots, if

$\displaystyle D\ge 0$

Þ    $\displaystyle 9{{k}^{2}}-144\ge 0$

Þ    $\displaystyle 9\,({{k}^{2}}-16)\ge 0$

Þ    $\displaystyle {{k}^{2}}-16\ge 0$            $\displaystyle [\because \,ab>0\,\,\text{and}\,\,a>0\Rightarrow b>0]$

Þ    $\displaystyle k\le -4\,\,\text{or}\,\,k\ge 4$        $\displaystyle [\because \,\,{{x}^{2}}-{{a}^{2}}\ge 0\,\,\Rightarrow \,\,x\le \,\,-a\,\,\text{or,}\,\,x\ge \,a]$

(iii)    The given equation is $\displaystyle 5{{x}^{2}}-kx+1=0$

Here,    $\displaystyle a=5,\,\,b=-k$ and c = 1

Þ    $\displaystyle D={{b}^{2}}-4ac={{(-k)}^{2}}-4\times 5\times 1={{k}^{2}}-20$

The given equation will have real roots, if

$\displaystyle D\ge 0$

Þ    $\displaystyle {{k}^{2}}-20\ge 0$

Þ    $\displaystyle k\le -\sqrt{20}\,\,\text{or},\,k\ge \sqrt{20}$        $\displaystyle [\because \,\,{{x}^{2}}-{{a}^{2}}\ge 0\Rightarrow x\le -a\,\,\text{or},\,\,x\ge a]$

### Problem:

Find the values of k for which the equation $\displaystyle {{x}^{2}}+5kx+16=0$ has no real roots.

### Solution:

The given equation is $\displaystyle {{x}^{2}}+5kx+16=0$

Comparing the given equation with $\displaystyle a{{x}^{2}}+bx+c=0,$ we have a = 1, b = 5k, c = 16

$\displaystyle D={{b}^{2}}-4ac={{(5k)}^{2}}-4\times 1\times 16=25{{k}^{2}}-64$

The given equation will have no real roots if D < 0

Þ    $\displaystyle 25{{k}^{2}}-64<0$

Þ    $\displaystyle 25\left( {{k}^{2}}-\frac{64}{25} \right)<0$

Þ    $\displaystyle {{k}^{2}}-\frac{64}{25}<0$         [If ab < 0 and a > 0, then b < 0]

Þ    $\displaystyle -\frac{8}{5}        [If $\displaystyle {{x}^{2}}-{{a}^{2}}<0,$ then –a < x < a]

### Problem:

If – 4 is a root of equation $\displaystyle {{x}^{2}}+px-4=0$ and the equation $\displaystyle {{x}^{2}}+px+q=0$ has equal roots, find the values of p and q.

### Solution:

Since – 4 is a root of $\displaystyle {{x}^{2}}+px-4=0,$ we have

$\displaystyle {{(-4)}^{2}}+p(-4)-4=0$    Þ    $\displaystyle 16-4p-4=0$

Þ    $\displaystyle 4p=12$    Þ    $\displaystyle p=3$                …(i)

Putting p = 3 in equation $\displaystyle {{x}^{2}}+px+q=0,$ we have

$\displaystyle {{x}^{2}}+3x+q=0$

Equation will have equal roots if D = 0 i.e. $\displaystyle {{b}^{2}}-4ac=0$

Þ    $\displaystyle {{p}^{2}}-4q=0$    Þ    $\displaystyle {{(3)}^{2}}-4q=0$         [Using (i)]

Þ    9 – 4q = 0    Þ    $\displaystyle q=\frac{9}{4}$

Hence, p = 3 and $\displaystyle q=\frac{9}{4}.$

### Problem:

If the roots of the equation $\displaystyle (a-b){{x}^{2}}+(b-c)x+(c-a)=0$ are equal, prove that $\displaystyle 2a=b+c.$

### Solution:

The given equation is $\displaystyle (a-b){{x}^{2}}+(b-c)x+(c-a)=0$

Comparing the given equation with $\displaystyle a{{x}^{2}}+bx+c=0,$

We have, $\displaystyle a=(a-b),\,\,b=(b-c)$ and $\displaystyle c=(c-a)$

$\displaystyle D={{b}^{2}}-4ac={{(b-c)}^{2}}-4\,(a-b)\,(c-a)$

For real and equal roots,    $\displaystyle D=0$ \$\displaystyle {{b}^{2}}-4ac=0$

Þ    $\displaystyle {{(b-c)}^{2}}-4(a-b)\,(c-a)=0$

Þ    $\displaystyle {{b}^{2}}+{{c}^{2}}-2bc-4ac+4bc+4{{a}^{2}}-4ab=0$

Þ    $\displaystyle 4{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-4ab+2bc-4ac=0$

Þ    $\displaystyle {{(-2a)}^{2}}+{{(b)}^{2}}+{{(c)}^{2}}+2(-2a)b+2(b)(c)+2c(-2a)=0$

Þ    $\displaystyle {{(-2a+b+c)}^{2}}=0$

or     $\displaystyle -2a+b+c=0$

or     $\displaystyle 2a=b+c.$

### Problem:

If – 5 is a root of the quadratic equation $\displaystyle 2{{x}^{2}}+px-15=0$ and the quadratic equation $\displaystyle p({{x}^{2}}+x)+k=0$ has equal roots, find the value of k.

### Solution:

Since – 5 is a root of the equation $\displaystyle 2{{x}^{2}}+px-15=0.$ Therefore,

$\displaystyle 2\,{{(-5)}^{2}}-5p-15=0$

Þ    $\displaystyle 50-5p-15=0$

Þ    $\displaystyle 5p=35$

Þ    $\displaystyle p=7$

Putting $\displaystyle p=7$ in $\displaystyle p({{x}^{2}}+x)+k=0,$ we get

$\displaystyle 7{{x}^{2}}+7x+k=0$

This equation will have equal roots, if

Discriminant = 0

Þ    49 – 4 ´ 7 ´ k = 0

Þ    $\displaystyle k=\frac{49}{28}\Rightarrow k=\frac{7}{4}$

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