What is the square root of 98 minus, square root of 24 plus the square root of 32?

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What is the square root of 98 minus, square root of 24 plus the square root of 32?

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Answer 1 · 2018-05-21T09:35:46

$11 \cdot \sqrt{2} - 2 \cdot \sqrt{6}$ $11*sqrt(2)-2*sqrt(6)$

Explanation:

$\sqrt{98} = \sqrt{2 \cdot 49} = \sqrt{2} \cdot 7$ $sqrt(98)=sqrt(2*49)=sqrt(2)*7$
$\sqrt{24} = \sqrt{6 \cdot 4} = 2 \sqrt{6}$ $sqrt(24)=sqrt(6*4)=2sqrt(6)$
$\sqrt{32} = \sqrt{2 \cdot 16} = 4 \cdot \sqrt{2}$ $sqrt(32)=sqrt(2*16)=4*sqrt(2)$

Answer 2 · 2018-05-21T09:40:17

$11 \sqrt{2} - 2 \sqrt{6}$ $11sqrt(2)-2sqrt(6)$

Explanation:

$\sqrt{98} = \sqrt{2 \times 7 \times 7} = 7 \sqrt{2}$ $sqrt(98) = sqrt(2xx7xx7)=7sqrt(2)$
$\sqrt{24} = \sqrt{2 \times 2 \times 2 \times 3} = 2 \sqrt{6}$ $sqrt(24) = sqrt(2xx2xx2xx3)=2sqrt(6)$
$\sqrt{32} = \sqrt{2 \times 2 \times 2 \times 2 \times 2} = 4 \sqrt{2}$ $sqrt(32) = sqrt(2xx2xx2xx2xx2)=4sqrt(2)$

$7 \sqrt{2} - 2 \sqrt{6} + 4 \sqrt{2} = 11 \sqrt{2} - 2 \sqrt{6}$ $7sqrt(2)-2sqrt(6)+4sqrt(2)=11sqrt(2)-2sqrt(6)$

Answer 3 · 2018-05-21T09:43:25

$11 \sqrt{2} - 2 \sqrt{6}$ $11sqrt2-2sqrt6$

Explanation:

$using the law of radicals$ $"using the "color(blue)"law of radicals"$

$∙ x \sqrt{a} \times \sqrt{b} \Leftrightarrow \sqrt{a b}$ $•color(white)(x)sqrtaxxsqrtbhArrsqrt(ab)$

$simplifying each radical gives$ $"simplifying each radical gives"$

$\sqrt{98} = \sqrt{49 \times 2} = \sqrt{49} \times \sqrt{2} = 7 \sqrt{2}$ $sqrt98=sqrt(49xx2)=sqrt49xxsqrt2=7sqrt2$

$\sqrt{24} = \sqrt{4 \times 6} = \sqrt{4} \times \sqrt{6} = 2 \sqrt{6}$ $sqrt24=sqrt(4xx6)=sqrt4xxsqrt6=2sqrt6$

$\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4 \sqrt{2}$ $sqrt32=sqrt(16xx2)=sqrt16xxsqrt2=4sqrt2$

$\Rightarrow \sqrt{98} - \sqrt{24} + \sqrt{32}$ $rArrsqrt98-sqrt24+sqrt32$

$= 7 \sqrt{2} - 2 \sqrt{6} + 4 \sqrt{2}$ $=color(blue)(7sqrt2)-2sqrt6color(blue)(+4sqrt2)$

$= 11 \sqrt{2} - 2 \sqrt{6}$ $=11sqrt2-2sqrt6$

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What is the square root of 98 minus, square root of 24 plus the square root of 32?

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