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}<k<\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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