Question

simplify the following
$\sqrt{98]$
$\sqrt{16 b^{5} }$
$3\sqrt{5m} x 4\sqrt{\frac{1}{5} m3}$
$\sqrt{26x2}$

Answer 1

Step-by-step explanation:

Simplify:

• $\sqrt{98} = \sqrt{49*2} = \sqrt{7^2*2} = 7\sqrt{2}$

• $\sqrt{16b^5} = \sqrt{4^2(b^2)^2b} = 4b^2\sqrt{b}$

• $3\sqrt{5mx^4} \sqrt{1/5m^3} = 3\sqrt{5*1/5*m^4x^4} = 3\sqrt{(m^2)^2(x^2)^2} = 3m^2x^2$

• $\sqrt{26x^2} = x\sqrt{26}$

Answer 2

Answer:

$→ \sqrt{98} = \sqrt{7 \times 7 \times2} = 7 \sqrt{2}✓$

$→ \sqrt{ 16{b}^{5} }= \sqrt{4 \times 4 \times {b}^{5} } = 4 {b}^{ \frac{5}{2} } ✓$

$→3 \sqrt{5 mx^{4} } \sqrt{ \frac{1}{5} {m}^{3}} = 3 \sqrt{5 \times \frac{1}{5} {m}^{4} {x}^{4} } = 3 {m}^{2} {x}^{2} ✓$

$→\sqrt{26 {x}^{2} } = \sqrt{26} x✓$