Question

x-2/x+2+x+2/x-2=4
solve it urgent!!​

Answer 1

$\large\sf\underline{Given\::}$

Equation :

• $\sf\:\frac{x-2}{x+2} + \frac{x+2}{x-2}=4$

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$\large\sf\underline{To\:find\::}$

• The value of x .

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$\large\sf\underline{Solution\::}$

$\sf\:\frac{x-2}{x+2} + \frac{x+2}{x-2}=4$

• LCM of (x + 2) (x - 2) = (x + 2) (x - 2)

$\sf\:\implies\:\frac{(x-2)(x-2)+(x+2)(x+2)}{(x+2)(x-2)}=4$

• Multiplying the terms

$\sf\:\implies\:\frac{(x-2)^{2}+(x+2)^{2}}{(x+2)(x-2)}=4$

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• Using algebraic identity in the numerator and denominator

Identity to be used in numerator :

• $\bf\:(a+b)^{2}=a^{2}+2ab+b^{2}$

• $\bf\:(a-b)^{2}=a^{2}-2ab+b^{2}$

Identity to be used in denominator :

• $\bf\:(a+b)(a-b)=a^{2}-b^{2}$

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$\sf\:\implies\:\frac{(x^{2}-2 \times x \times 2 + 2^{2})+(x^{2}+2 \times x \times 2 + 2^{2})}{x^{2}-2^{2}}=4$

$\sf\:\implies\:\frac{(x^{2}-4x + 4)+(x^{2}+4x+4)}{x^{2}-4}=4$

• Removing the brackets

$\sf\:\implies\:\frac{x^{2}-4x + 4+x^{2}+4x+4}{x^{2}-4}=4$

• Arranging the terms and proceeding with simple calculations

$\sf\:\implies\:\frac{x^{2}+x^{2}-4x+4x+4+4}{x^{2}-4}=4$

$\sf\:\implies\:\frac{2x^{2}-\cancel{4x}+\cancel{4x}+8}{x^{2}-4}=4$

$\sf\:\implies\:\frac{2x^{2}+8}{x^{2}-4}=4$

• Cross multiplying

$\sf\:\implies\:4(x^{2}-4) = 2x^{2}+8$

• Multiplying the terms in LHS

$\sf\:\implies\:4x^{2}-16 = 2x^{2}+8$

• Transposing $\sf\:+2x^{2}$ from RHS to LHS it becomes $\sf\:-2x^{2}$ .

$\sf\:\implies\:4x^{2}- 2x^{2}-16=8$

$\sf\:\implies\: 2x^{2}-16=8$

• Transposing -16 to RHS it becomes +16

$\sf\:\implies\: 2x^{2}=8+16$

$\sf\:\implies\: 2x^{2}=24$

• Transposing 2 to other side it goes to denominator

$\sf\:\implies\: x^{2}=\frac{24}{2}$

• Reducing the fraction to lower terms

$\sf\:\implies\: x^{2}=\cancel{\frac{24}{2}}$

$\sf\:\implies\: x^{2}=12$

• Transposing square to other side it becomes √

$\small{\underline{\boxed{\mathrm\red{:\implies\:x\:=\:\sqrt{12}}}}}$ ★

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$\large\sf\underline{Verifying\::}$

Not sure about the answer !? Let's verify it.

In order to verify we would simply plug the value of x as √12 in the given equation. Doing so if we get LHS equal to RHS , our answer must be correct.

Equation :

$\sf\:\frac{x-2}{x+2} + \frac{x+2}{x-2}=4$

• Plugging the value of x as √12

$\sf\to\:\frac{\sqrt{12}-2}{\sqrt{12}+2} + \frac{\sqrt{12}+2}{\sqrt{12}-2}=4$

• LCM of (√12 + 2) (√12 - 2) = (√12 + 2) (√12 - 2)

$\sf\to\:\frac{(\sqrt{12}-2)^{2}+(\sqrt{12}+2)^{2}}{(\sqrt{12}^{2}-2^{2}}=4$

$\sf\to\:\frac{[(\sqrt{12})^{2}-2 \times \sqrt{12} \times 2 + 2^{2}]+[(\sqrt{12})^{2}+2 \times \sqrt{12} \times 2 + 2^{2}]}{12-4}=4$

$\sf\to\:\frac{[12-4 \sqrt{12} +4]+[12+4 \sqrt{12}+ 4]}{8}=4$

$\sf\to\:\frac{12-4 \sqrt{12} +4+12+4 \sqrt{12}+ 4}{8}=4$

$\sf\to\:\frac{12+12+4+4-\cancel{4 \sqrt{12}} + \cancel{4 \sqrt{12}}}{8}=4$

$\sf\to\:\frac{32}{8}=4$

• Cross multiplying

$\sf\to\:32=4 \times 8$

$\sf\to\:32=32$

$\bf\to\:LHS\:=\:RHS$

$\small{\mathfrak\blue{hence\:verified\:!!}}$

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!! Hope it helps !!