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How do you simplify $\sqrt{5x-4}-\sqrt{x+8}=2$ ?

Answer
VerifiedVerified
550.2k+ views
Hint: We can solve this question by squaring both sides. After squaring both sides there will be another term which will be under square root. We can separate that term and then we will square both sides. The equation will be a quadratic equation, so there will be 2 values of x.

Complete step by step answer:
The given equation is $\sqrt{5x-4}-\sqrt{x+8}=2$
Squaring both sides we get
$6x+4-2\sqrt{5{{x}^{2}}+36x-32}=4$
We can see another term inside the square root so we will send that term to RHS and rest of the term to LHS
 $6x=2\sqrt{5{{x}^{2}}+36x-32}$
Now again squaring both sides we get
 $36{{x}^{2}}=20{{x}^{2}}+144x-128$
Further solving we get
 $16{{x}^{2}}-144x+128=0$
Taking 16 common from the equation we get
${{x}^{2}}-9x+8=0$
By factorization of ${{x}^{2}}+9x-8=0$ we get ( x – 1 ) ( x – 8 ) = 0
So x is equal to 1 or 8.
But if we put x equal to 1 in $\sqrt{5x-4}-\sqrt{x+8}$ , we get $\sqrt{5\times 1-4}-\sqrt{1+8}$
$\Rightarrow \sqrt{5\times 1-4}-\sqrt{1+8}=\sqrt{1}-\sqrt{9}$
$\Rightarrow \sqrt{5\times 1-4}-\sqrt{1+8}=1-3$
$\Rightarrow \sqrt{5\times 1-4}-\sqrt{1+8}=-2$
So -1 is not the correct answer.
If we put x equal to 8
$\Rightarrow \sqrt{5\times 8-4}-\sqrt{8+8}=\sqrt{36}-\sqrt{16}$
$\Rightarrow \sqrt{5\times 1-4}-\sqrt{1+8}=6-4$
$\Rightarrow \sqrt{5\times 1-4}-\sqrt{1+8}=2$
So 8 is the correct answer.

Note:
While solving any question by squaring both sides, always check the answer by putting the value in the equation. For example, we have to calculate $\sqrt{9}$. We will take it as x and squaring both sides we get ${{x}^{2}}$ equal to 9. Solution of ${{x}^{2}}$ = 9 is 3 and – 3. But the correct value of $\sqrt{9}$ is 3, - 3 is not the correct answer.
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