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ICSE Class 10 Mathematics Question 18 of 21

Solved 2023 Question Paper ICSE Class 10 Mathematics — Question 16

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Question 9(i)

Using componendo and dividendo solve for x : 2x+2+2x−12x+2−2x−1=3\dfrac{\sqrt{2x + 2} + \sqrt{2x - 1}}{\sqrt{2x + 2} - \sqrt{2x - 1}} = 3.

Answer

Given : 2x+2+2x−12x+2−2x−1=3\dfrac{\sqrt{2x + 2} + \sqrt{2x - 1}}{\sqrt{2x + 2} - \sqrt{2x - 1}} = 3

Applying componendo and dividendo, we get :

⇒2x+2+2x−1+2x+2−2x−12x+2+2x−1−(2x+2−2x−1)=3+13−1⇒22x+222x−1=42⇒2x+22x−1=2⇒2x+2=22x−1\Rightarrow \dfrac{\sqrt{2x + 2} + \sqrt{2x - 1} + \sqrt{2x + 2} - \sqrt{2x - 1}}{\sqrt{2x + 2} + \sqrt{2x - 1} - (\sqrt{2x + 2} - \sqrt{2x - 1})} = \dfrac{3 + 1}{3 - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{2x + 2}}{2\sqrt{2x - 1}} = \dfrac{4}{2} \\[1em] \Rightarrow \dfrac{\sqrt{2x + 2}}{\sqrt{2x - 1}} = 2 \\[1em] \Rightarrow \sqrt{2x + 2} = 2\sqrt{2x - 1}

Squaring both sides we get :

⇒ 2x + 2 = 4(2x - 1)

⇒ 2x + 2 = 8x - 4

⇒ 8x - 2x = 2 + 4

⇒ 6x = 6

⇒ x = 66\dfrac{6}{6} = 1.

Hence, x = 1.