Question

Solve:$\sqrt{4-x}+\sqrt{x+9}=5$

• A. $x=0$
• B. $x=1$
• C. $x=2$
• D. $x=3$

Answer 1

The correct option is A $x=0$

$\sqrt{4-x}+\sqrt{x+9}=5$

Squaring both sides, we have

$(4-x)+(x+9)+2\sqrt{36-x^2-5x}=25$

$13+2\sqrt{36-x^2-5x}=25$

$2\sqrt{36-x^2-5x}=25-13$

$\sqrt{36-x^2-5x}=6$

Squaring again both sides, we have

$36-x^2-5x=36$

$x^2+5x=0$

$\Rightarrow x=0,-5$

But $x<4$

Therefore, $x=0$