Question

Find the integral value of $x$ such that $\sqrt{4x+1}+\sqrt{7-x}=6$

Answer 1

$\sqrt{4x+1}+\sqrt{7-x}=6$

$⟹\sqrt{4x+1}=6-\sqrt{7-x}$

Squaring both sides,

$⟹4x+1=(6-\sqrt{7-x}{)}^2=36+(7-x)-12\sqrt{7-x}$

$⟹12\sqrt{7-x}=36+7-x-4x-1=42-5x$

Again Squaring both sides

$⟹\left(144\right)(\sqrt{7-x}{)}^2=(42-5x{)}^2$

$⟹144(7-x)=1764+25x^2-420x$

$⟹25x^2-276x+756=0$

Using quadratic formula,

$x=\frac{+276\pm \sqrt{(276{)}^2-4(25\left)\right(756)}}{50}$

$x=\frac{276\pm \sqrt{76176-75600}}{50}=\frac{276\pm 24}{50}$

$x=\frac{300}{50}=6$ or $x=\frac{252}{50}=\frac{126}{25}$

Values of $x=6,\frac{126}{25}$

$\therefore x=6$