Question

If the difference of roots of the equation $x^2-px+q=0$ is $1$ then :

• A. $p^2+4q^2={\left(1+2q\right)}^2$
• B. $4p^2+q^2={\left(1+2q\right)}^2$
• C. $p^2-4q^2={\left(1+2q\right)}^2$
• D. $4p^2+q^2={\left(1-2q\right)}^2$

Answer 1

The correct option is A

$p^2+4q^2={\left(1+2q\right)}^2$

Explanation for the correct option.

Step 1: Relation between roots

Given an equation, $x^2-px+q=0$ and difference of roots is $1$.

Let the root of the quadratic equation be $\alpha ,\beta$
$\therefore \alpha -\beta =1$

Squaring both sides we get:
$\begin{array}{rcl}(\alpha -\beta {)}^2& =& 1\\ {\alpha }^2+{\beta }^2-2\alpha \beta & =& 1\\ (\alpha +\beta {)}^2-2\alpha \beta -2\alpha \beta & =& 1\\ (\alpha +\beta {)}^2-4\alpha \beta & =& 1....\left(1\right)\end{array}$
The sum of roots is given by:

$\alpha +\beta =p$
The product of roots is given by:

$\alpha \beta =q$

Step 2: Find the value of expression
Substituting these values in equation (1), we get:
$$\begin{gathered} p^2-4q=1 \\ \Rightarrow {\mathrm{p}}^2=1+4\mathrm{q} \end{gathered}$$

This gives:
$\begin{array}{rcl}{\mathrm{p}}^2+4{\mathrm{q}}^2& =& 1+4\mathrm{q}+4{\mathrm{q}}^2\\ & =& 4{\mathrm{q}}^2+4\mathrm{q}+1\\ {\mathrm{p}}^2+4{\mathrm{q}}^2& =& (2\mathrm{q}+1{)}^2\end{array}$
Hence, option A is correct.