# Can You Solve this Algebra Math Problem

# Find the value of x, when the square root of 11 minus x, minus square root of x plus 6 equal to 3

\sqrt{11-x}-\sqrt{6+x}=3

x=?

## Solution to this Algebra Math Problem

\sqrt{11-x}-\sqrt{6+x}=3

Square both sides, then

(\sqrt{11-x}-\sqrt{6+x})^2=3^2

\Rightarrow 11-x-2\sqrt{(11-x)(6+x)}+6+x=9

\Rightarrow \space -2\sqrt{(11-x)(6+x)}+8=0

\Rightarrow \space \sqrt{(11-x)(6+x)}=4

again square both sides

(11-x)(6+x)=4^2=16

\Rightarrow \space 66+11x-6x-x^2=16

\Rightarrow \space x^2-5x-50=0

## This is a quadratic equation, so

x= \dfrac{5\pm\sqrt{(-5)^2-4\times 1\times(-50)}}{2\times1}

\Rightarrow \space x= \dfrac{5\pm\sqrt{(-5)^2-4\times 1\times(-50)}}{2\times1}

\Rightarrow \space x= \dfrac{5\pm\sqrt{25+200}}{2}

\Rightarrow \space x= \dfrac{5\pm15}{2}

\Rightarrow \space x= 10 \space \space \& \space \space x=-5

### Now we can check the solution is right or wrong

When *x* = 10, then

\sqrt{11-x}-\sqrt{6+x}\stackrel{?}{=}3

\Rightarrow \space \sqrt{11-10}-\sqrt{6+10}\stackrel{?}{=}3

\Rightarrow \space 1-4\stackrel{?}{=}3

\Rightarrow \space -3\neq3

So ** x = 10 ** is

**not a solution**to this problem

When *x* = -5, then

\sqrt{11-x}-\sqrt{6+x}\stackrel{?}{=}3

\Rightarrow \space \sqrt{11-(-5)}-\sqrt{6+(-5)}\stackrel{?}{=}3

\Rightarrow \space \sqrt{16}-\sqrt{1}\stackrel{?}{=}3

\Rightarrow \space 4-1\stackrel{?}{=}3

\Rightarrow \space 3=3

So ** x = -5 ** is

**a solution**to this problem

Finally,** **we got ** x = -5 is the only solution **to this algebra math problem