Question

Find the value of p, if the following quadratic equation has equal roots
${4x}^{2} - (p - 2)x + 1 = 0$

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Answer 1

GIVEN :–

$\\ \: \: \to \: \: \: \rm A\: \: quadratic \: \: equation \: \: 4x{}^{2} - (p - 2)x + 1 = 0 \: \: have \: \: equal \: \: roots.$

TO FIND :–

• Value of 'p' = ?

SOLUTION :–

• If a quadratic equation ax² + bx + c = 0 has equal roots , then Discriminant of quadratic equation will be zero.

$\\ \: \: \to \: \: \ \rm \large \: D = 0$

• And we know that –

$\\ \: \: \to \: \: \ \rm \large \: D = {b}^{2} - 4ac$

• Here –

$\\ \: \: \: \: \: \to \: \: \ \rm \large a = 4$

$\\ \: \: \: \: \: \to \: \: \ \rm \large b = - (p - 2)$

$\\ \: \: \: \: \: \to \: \: \ \rm \large c = 1$

• So that –

$\\ \: \implies \: \ \rm {b}^{2} - 4ac = 0$

$\\ \: \implies \: \ \rm {( - (p - 2))}^{2} - 4(4)(1) = 0$

$\\ \: \implies \: \ \rm { (p - 2)}^{2} - 16 = 0$

$\\ \: \implies \: \ \rm { (p - 2)}^{2} = 16$

$\\ \: \implies \: \ \rm p - 2 = \sqrt{16}$

$\\ \: \implies \: \ \rm p - 2 = \pm \: 4$

Take positive(+) sign :–

$\\ \: \implies \: \ \rm p - 2 = \: 4$

$\\ \: \implies \: \ \rm p = \: 4 + 2$

$\\ \: \implies \: \large \rm p = \: 6$

Take negative(-) sign :–

$\\ \: \implies \: \ \rm p - 2 = \: -4$

$\\ \: \implies \: \ \rm p = \: -4 + 2$

$\\ \: \implies \: \large \rm p = \: -2$