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

May 21, 2018

$11 \cdot \sqrt{2} - 2 \cdot \sqrt{6}$

#### Explanation:

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

May 21, 2018

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

#### Explanation:

$\sqrt{98} = \sqrt{2 \times 7 \times 7} = 7 \sqrt{2}$
$\sqrt{24} = \sqrt{2 \times 2 \times 2 \times 3} = 2 \sqrt{6}$
$\sqrt{32} = \sqrt{2 \times 2 \times 2 \times 2 \times 2} = 4 \sqrt{2}$

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

May 21, 2018

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

#### Explanation:

$\text{using the "color(blue)"law of radicals}$

•color(white)(x)sqrtaxxsqrtbhArrsqrt(ab)

$\text{simplifying each radical gives}$

$\sqrt{98} = \sqrt{49 \times 2} = \sqrt{49} \times \sqrt{2} = 7 \sqrt{2}$

$\sqrt{24} = \sqrt{4 \times 6} = \sqrt{4} \times \sqrt{6} = 2 \sqrt{6}$

$\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4 \sqrt{2}$

$\Rightarrow \sqrt{98} - \sqrt{24} + \sqrt{32}$

$= \textcolor{b l u e}{7 \sqrt{2}} - 2 \sqrt{6} \textcolor{b l u e}{+ 4 \sqrt{2}}$

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