Nature Of Roots Depending Upon Coefficients And Discriminant - Practice Questions & MCQ

Let the quadratic equation is $a x^2+b x+c=0,(a, b, c \in R)$

$D$ (called the discriminant of the equation) $=b^2-4 a d$

The roots of this equation are given by

$x_1=\frac{-b+\sqrt{D}}{2 a}$ and $x_2=\frac{-b-\sqrt{D}}{2 a}$

i) if D < 0, then both roots are non-real (imaginary numbers), and the roots will be conjugate of each other, which means if p + iq is one of  the roots then the other root will be p - iq

ii) If D > 0, then roots will be real and distinc

iii) D = 0, then roots will be real and equal, and they equal$\mathrm{x}_1=\mathrm{x}_2=\frac{-\mathrm{b}}{2 \mathrm{a}}$

Special cases of case ii (D > 0)

i) if a,b,c are rational numbers (Q) and

  If D is a perfect square, then roots are rational

  If D is not a perfect square then roots are irrational (in this case if$p+\sqrt{q}$ is one root of the quadratic equation then another root will be $p-\sqrt{q}$ )

ii) If a = 1 and b and c are integers and  

  If D is a perfect square, then roots are integers

 If D is not a perfect square then roots are non-integer values

Let alpha and beta be two roots of a quadratic equation. So, we have

$\begin{aligned} & \alpha=\frac{-b-\sqrt{D}}{2 a} \\ & \beta=\frac{-b+\sqrt{D}}{2 a}\end{aligned}$

The sum of roots:

$\alpha+\beta=\frac{-\mathrm{b}-\sqrt{\mathrm{D}}}{2\mathrm{a}}+\frac{\mathrm{b}+\sqrt{\mathrm{D}}}{2 \mathrm{a}}=\frac{-\mathrm{b}}{\mathrm{a}}$

Product of roots:

$\begin{aligned} & \alpha \cdot \beta=\left(\frac{-\mathrm{b}-\sqrt{\mathrm{D}}}{2 \mathrm{a}}\right) \cdot\left(\frac{-\mathrm{b}+\sqrt{\mathrm{D}}}{2\mathrm{a}}\right)\\&=\frac{\mathrm{b}^2-\mathrm{D}}{4\mathrm{a}^2}=\frac{\mathrm{b}^2-\mathrm{b}^2+4\mathrm{ac}}{4\mathrm{a}^2}=\frac{4 \mathrm{ac}}{4 \mathrm{a}^2}=\frac{\mathrm{c}}{\mathrm{a}}\end{aligned}$

The difference of root can also be found in the same way by manipulating the terms

$\alpha-\beta=\left|\frac{\sqrt{D}}{a}\right|$

Important Results

(i) $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta$
(ii) $\alpha^2-\beta^2=(\alpha+\beta)(\alpha-\beta)$
(iii) $\alpha^3+\beta^3=(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)$
(iv) $\alpha^3-\beta^3=(\alpha-\beta)^3+3 \alpha \beta(\alpha-\beta)$