Question

Solve the following inequality:
$\sqrt{3x^2+5x+7}-\sqrt{3x^2+5x+2}>1.$

Answer 1

$\sqrt{3x^{+}5x+7}-\sqrt{3x^2+5x+2}>1$-Equation $1$

$\Rightarrow$ For definition of sqaure roots,

$3x^2+5x+7\ge 0$-Equation 2 $\forall x\in R(asa>0)(D<0)$

$3x^2+5x+2\ge 0$-Equation $3$

$(x+1)(x+\frac{2}{3})\ge 0$

From Equation $1$,

$\sqrt{3x^2+5x+7}>1+\sqrt{3x^2+5x+2}$

and squaring on both sides,$\sqrt{3x^2+5x+7}>1+2\sqrt{3x^2+5x+2}+3x^2+5x+2$

$\Rightarrow 5>1+2\sqrt{3x^2+5x+2}$

$\Rightarrow 2>\sqrt{3x^2+5x+2}$

Again on squaring, $4>3x^2+5x+2$

$\Rightarrow (x-\frac{1}{3})\left)\right(x+2)<0$-Equation 4

From Equation$2$, $3$ and $4$

$x\in (-2,-1]\cup [-\frac{2}{3},\frac{1}{3})$