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