Question

$\sqrt{{x}^{2} - 10x + 24 } > 0$
Find the value of x. Sets question​

Answer 1

Solution!!

$\sf \sqrt{x^{2}-10x+24}>0$

Determine the defined range.

$\sf \sqrt{x^{2}-10x+24}>0,\:x\in \left<-\infty ,4\right]\cup \left[6, +\infty \right>$

Since, the left-hand side is always positive or 0, the statement is true for any value of x, except when $\sf \sqrt{x^{2}-10x+24}=0$.

$\sf$

$\sf \sqrt{x^{2}-10x+24}=0$

$\sf x^{2}-10x+24=0$

$\sf x^{2}-4x-6x+24=0$

$\sf x(x-4)-6(x-4)=0$

$\sf (x-4)(x-6)=0$

When the product of factors equal 0, atleast one factor is 0.

$\sf x-4=0$

$\sf x-6=0$

$\sf x=4$

$\sf x=6$

Hence, the inequality is true for any value of x, except when x = 4, x = 6.

$\sf x\in R\ \left\{4,6\right\},\:x\in \left<-\infty ,4\right]\cup \left[6, +\infty \right>$

$\sf x\in \left<-\infty ,4\right>\cup \left<6, +\infty \right>$