Question

(√32+√48)/(√8+√12) answer​

Answer 1

Given :

• $\large\sf\:\frac{\sqrt{32}~+~\sqrt{48}}{\sqrt{8}~+~\sqrt{12}}$

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To :

• Simplify the given expression.

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Concept :

For this question we would split the bigger numbers under the square root into smaller numbers. Then expressing the smaller numbers in their square form, we could cancel the square root and the square of the numbers. Doing so and proceeding with simple calculations, we would get our answers.

Hope am clear ! Let's proceed :D

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Solution :

$\large\sf\:\frac{\sqrt{32}~+~\sqrt{48}}{\sqrt{8}~+~\sqrt{12}}$

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Splitting 32 , 48 , 8 and 12 :

• 32 = 4 × 4 × 2
• 48 = 4 × 4 × 3
• 8 = 2 × 2 × 2
• 12 = 2 × 2 × 3

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$\large\sf\implies\:\frac{\sqrt{4 \times 4 \times 2}~+~\sqrt{4 \times 4 \times 3}}{\sqrt{2 \times 2 \times 2}~+~\sqrt{2 \times 2 \times 3}}$

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$\large\sf\implies\:\frac{\sqrt{(4) ^{2} \times 2}~+~\sqrt{(4) ^{2} \times 3}}{\sqrt{(2)^{2} \times 2}~+~\sqrt{2^{2} \times 3}}$

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Taking the numbers with square out of the square root

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$\large\sf\:\implies\:\frac{4\sqrt{2}~+~4\sqrt{3}}{2\sqrt{2}~+~2\sqrt{3}}$

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Taking 4 as common from the numerator

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$\large\sf\:\implies\:\frac{4(\sqrt{2}~+~\sqrt{3})}{2\sqrt{2}~+~2\sqrt{3}}$

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Taking 2 as common from the denominator

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$\large\sf\:\implies\:\frac{4(\sqrt{2}~+~\sqrt{3})}{2(\sqrt{2}~+~\sqrt{3})}$

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Observing properly we notice ( √2+√3 ) can be cancelled from both numerator and denominator

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$\large\sf\:\implies\:\frac{4~~~\cancel{(\sqrt{2}~+~\sqrt{3})}}{2~~~\cancel{(\sqrt{2}~+~\sqrt{3})}}$

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$\large\sf\:\implies\:\frac{4}{2}$

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Reducing the fraction to lower terms

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$\large\sf\:\implies\:\cancel{\frac{4}{2}}$

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$\small{\underline{\boxed{\mathrm\purple{\implies\:2}}}}$ $\bf\color{red}{⋆}$

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‎Therefore simplified version of the given expression is $\large\mathrm\color{teal}{2}$ .

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