Question

x²+x-8
$\leqslant 6$
​

Answer 1

Answer

• x ≤ ( √57 - 1 ) / 2

Given

• x² + x - 8 ≤ 6

To Find

• x value

Concept Used

• Completing the square

Solution

$\rm x^2+x-8 \leq 6\\\\\implies \rm x^2+x\leq 6+8\\\\\implies \rm x^2+x \leq 14\\\\\implies \rm x^2+2.\dfrac{1}{2}.x \leq 14\\\\\implies \rm x^2+2.\dfrac{1}{2}.x+\bigg(\dfrac{1}{2}\bigg)^2 \leq 14+\bigg(\dfrac{1}{2}\bigg)^2 \\\\ \rm [Since\ ,Adding\ (1/2)^2\ on\ both\ sides]\\\\\implies \rm \bigg(x+\dfrac{1}{2}\bigg)^2 \leq 14+\dfrac{1}{4}\\\\\implies \rm \bigg(x+\dfrac{1}{2}\bigg)^2 \leq \dfrac{57}{4}\\\\\implies \rm x+\dfrac{1}{2} \leq \dfrac{\sqrt{57}}{2}$

$\implies \rm x \leq \dfrac{\sqrt{57}}{2}-\dfrac{1}{2}\\\\\implies \bf x \leq \dfrac{(\sqrt{57}-1)}{2}\ \;\; \bigstar$