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CAT 2024 Slot 2 — QA Question 17

Square root of SurdsEasy

If (x+62)12 - (x-62)12 = 22, then x equals

Answer & solution

Answer: 11

Solution

Medium

Square both sides to remove the outer square roots. The cross term gives a single inner radical; isolate and square once more to solve the resulting linear equation. Check the root.

1

Square the given equation. Let L=x+62−x−62=22L=\sqrt{x+6\sqrt2}-\sqrt{x-6\sqrt2}=2\sqrt2.

(x+62)+(x−62)−2(x+62)(x−62)=(22)2⇒ 2x−2x2−72=8((a−b)(a+b)=x2−72)⇒ x−x2−72=4\begin{aligned} &(x+6\sqrt2)+(x-6\sqrt2)-2\sqrt{(x+6\sqrt2)(x-6\sqrt2)}=(2\sqrt2)^2\\ &\Rightarrow\ 2x-2\sqrt{x^2-72}=8 \quad\text{(}(a-b)(a+b)=x^2-72\text{)}\\ &\Rightarrow\ x-\sqrt{x^2-72}=4 \end{aligned}
2

Isolate and square again.

x2−72=x−4⇒ x2−72=x2−8x+16(square both sides)⇒ 8x=88 ⇒ x=11\begin{aligned} &\sqrt{x^2-72}=x-4\\ &\Rightarrow\ x^2-72=x^2-8x+16 \quad\text{(square both sides)}\\ &\Rightarrow\ 8x=88\ \Rightarrow\ x=11 \end{aligned}
3

Check. x=11≥4x=11\ge4 keeps x−4≥0x-4\ge0, so squaring introduced no false root; also x2−72=49>0x^2-72=49>0.

11+62−11−62=(3+2)2−(3−2)2=22 ✓\begin{aligned} &\sqrt{11+6\sqrt2}-\sqrt{11-6\sqrt2}=\sqrt{(3+\sqrt2)^2}-\sqrt{(3-\sqrt2)^2}=2\sqrt2\ \checkmark \end{aligned}
x=11x=11
CAT 2024 Slot 2 QA Q17: If x + 6 2 1 2 - x - 6 2 1 2 = 2 2 , then x equals — Solution | TheCATExam