CMAT 2016 Solved Question Paper

(b) Since, x = 1 is a root of equation.
${\mathit{x}}^2$ + kx - 4 = 0
⇒ $1^2$ + k × 1 - 4 = 0 ⇒ k = 3
Equation ${\mathit{x}}^2$ + kx + ${\mathit{k}}_2$ = 0 has equal roots.
By Sridharacharya method,
For equal roots, ${\mathit{b}}^2$ - 4ac = 0
⇒ ${\mathit{k}}^2$ - 4 × 1 + ${\mathit{k}}_2$ = 0
⇒ $3^2$ -4${\mathit{k}}_2$ = 0 $$[∵ k = 3]
⇒ 9 - 4${\mathit{k}}_2$ =0 ⇒ ${\mathit{k}}_2=\frac{9}{4}$
∴ Required equation, ${\mathit{x}}^2-\left(\mathit{k}+{\mathit{k}}_2\right)\mathit{x}+\mathit{k}{\mathit{k}}_2$ = 0
⇒ ${\mathit{x}}^2-\left(\frac{9}{4}+3\right)\mathit{x}+\frac{9}{4}\times 3$ = 0
⇒ 4${\mathit{x}}^2$ - 21x + 27 = 0