Question 134

If $$\ \sqrt{7x+12}+\sqrt{7x-12}=3+\sqrt{33}\ $$, then what is the value of x?

Solution

Expression : $$\ \sqrt{7x+12}+\sqrt{7x-12}=3+\sqrt{33}\ $$

$$(7x-12)$$ must be greater than equal to 0.

=> $$7x-12\geq0$$

=> $$x\geq\frac{12}{7}$$

Thus, first two options are eliminated. Putting $$x=3$$ in above equation,

=> $$\sqrt{7(3)+12}+\sqrt{7(3)-12}$$

=> $$\sqrt{33}+\sqrt9$$

= $$3+\sqrt{33}=$$ R.H.S.

=> Ans - (C)

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