How do you subtract $\sqrt{8 x^{3}} - 4 x \sqrt{98 x}$ $\sqrt{8x^3}-4x\sqrt{98x}$ ?

Question

8941 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-13T20:33:15

$\sqrt{8 x^{3}} - 4 x \sqrt{98 x}$ $sqrt(8x^3)−4xsqrt(98x)$

Symplify the expreessions-:

$\sqrt{2 x {(2 x)}^{2}} - 4 x \sqrt{2 x 7^{2}}$ $sqrt(2x(2x)^2)-4xsqrt(2x7^2$

Applying the rule $\sqrt{a^{2}} = a$ $sqrt(a^2)=a$ :

$2 x \sqrt{2 x} - 7 \times 4 x \sqrt{2 x}$ $2xsqrt(2x)-7xx4xsqrt(2x)$

$(2 x - 28 x) \sqrt{2 x}$ $(2x-28x)sqrt(2x)$

$- 26 x \sqrt{2 x}$ $-26xsqrt(2x)$

How do you subtract √8x3−4x√98x\sqrt{8x^3}-4x\sqrt{98x}?