How do you simplify $\sqrt{28 x^{3} y^{4}}$ $sqrt(28x^3y^4)$ ?

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How do you simplify $\sqrt{28 x^{3} y^{4}}$ $sqrt(28x^3y^4)$ ?

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Answer 1 · 2015-04-18T19:50:41

$\sqrt{28 x^{3} y^{4}}$ $sqrt(28x^3y^4)$

$= \sqrt{(7 \cdot 4) (x^{2} \cdot x) {(y^{2})}^{2}}$ $= sqrt((7*4)(x^2*x)(y^2)^2)$

$= \sqrt{(7 \cdot 2^{2}) (x^{2} \cdot x) {(y^{2})}^{2}}$ $= sqrt((7*2^2)(x^2*x)(y^2)^2)$

$= 2 \cdot x \cdot y^{2} \sqrt{7 x}$ $color(green)( = 2*x*y^2sqrt(7x)$ is the Simplified Form of $\sqrt{28 x^{3} y^{4}}$ $sqrt(28x^3y^4)$

Answer 2 · 2015-09-08T18:28:32

Simplifying the values can work out here.

Explanation:

the above can also be written as $\sqrt{28} {\sqrt{x}}^{3} {\sqrt{y}}^{4}$ $sqrt28sqrtx^3sqrty^4$ for convenience.
1) $\sqrt{28}$ $sqrt28$ =2 $\sqrt{7}$ $sqrt7$
2) ${\sqrt{x}}^{3}$ $sqrtx^3$ =x $\sqrt{x}$ $sqrtx$
3) ${\sqrt{y}}^{4}$ $sqrty^4$ = $y^{2}$ $y^2$
Hence on using 1,2 and 3 we get the answer=2x $y^{2}$ $y^2$ $\sqrt{7}$ $sqrt7$ $\sqrt{x}$ $sqrtx$
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How do you simplify $\sqrt{28 x^{3} y^{4}}$ $sqrt(28x^3y^4)$ ?

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