Question

Match the column A with column B

Column A	Column B
(a) $\Delta =0$	(p) roots are rational
(b) $\Delta <0$	(q) roots are real and equal
(c) $\Delta >0$ and perfect square	(r) roots are irrational
(d) $\Delta >0$ and not perfect square	(s) roots are complex conjugates

(a) $a\to q;b\to s;c\to r;d\to p$
(b) $a\to s;b\to q;c\to p;d\to r$
(c) $a\to q;b\to s;c\to p;d\to r$
(d) None of these

Answer 1

Hint: We start solving the problem by recalling the definition of discriminant and roots of the quadratic equation $a{{x}^{2}}+bx+c=0$. We then assume each condition of the discriminant given in column A and substitute in the roots of the equation to verify the properties of the roots obtained. We then match them using these properties of roots.

Complete step-by-step answer:
According to the problem, we need to match the columns in the given column.

Column A	Column B
(a) $\Delta =0$	(p) roots are rational
(b) $\Delta <0$	(q) roots are real and equal
(c) $\Delta >0$ and perfect square	(r) roots are irrational
(d) $\Delta >0$ and not perfect square	(s) roots are complex conjugates

Let us recall the definition of discriminant of the given quadratic equation $a{{x}^{2}}+bx+c=0$.
We know that the discriminant of the quadratic equation $a{{x}^{2}}+bx+c=0$ is defined as $\Delta ={{b}^{2}}-4ac$ ---(1) and the roots of this quadratic equation is $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ ---(2).
Let us assume $\Delta =0$, which gives us ${{b}^{2}}-4ac=0$ from equation (1). Let us substitute this equation (2).
$\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-b\pm \sqrt{0}}{2a}=\dfrac{-b}{2a}$, which is only root (i.e., roots are real and equal).
So, ‘a’ is matching with ‘q’ $\left( a\to q \right)$ ---(3).
Let us assume $\Delta <0$, which gives us ${{b}^{2}}-4ac<0$ from equation (1) and assume it as $-d$. Let us substitute this equation (2).
$\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-b\pm \sqrt{-d}}{2a}=\dfrac{-b\pm i\sqrt{d}}{2a}$, which means complex conjugate root (i.e., roots are complex conjugates).
So, ‘b’ is matching with ‘s’ $\left( b\to s \right)$ ---(4).
Let us assume $\Delta >0$ and perfect square, which gives us ${{b}^{2}}-4ac>0$ from equation (1) and assume it as ${{s}^{2}}$. Let us substitute this equation (2).
$\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-b\pm \sqrt{{{s}^{2}}}}{2a}=\dfrac{-b\pm s}{2a}$, which means real and rational roots (i.e., roots are real and rational).
So, ‘c’ is matching with ‘p’ $\left( c\to p \right)$ ---(5).
Let us assume $\Delta >0$ and not a perfect square, which gives us ${{b}^{2}}-4ac>0$ from equation (1) and assume it as $s$. Let us substitute this equation (2).
$\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-b\pm \sqrt{s}}{2a}$, which means real and irrational roots (i.e., roots are real and irrational).
So, ‘d’ is matching with ‘r’ $\left( d\to r \right)$ ---(6).
From equations (3), (4), (5) and (6), we got $a\to q;b\to s;c\to p;d\to r$.

So, the correct answer is “Option (c)”.

Note: We should know the difference between the complex, rational and irrational numbers before solving this problem. We should know that the $\Delta $ present in the problem is discriminant of quadratic equations not the determinant of matrices. We should know that these are standard properties for the roots of quadratic equations. Similarly, we can expect problems to find the discriminant and properties of different quadratic equations.