If $a^{2} + b^{2} = z and a b = y$ $a^2 + b^2 = z and ab =y$ , what will be the equivalent of $4 z + 8 y$ $4z+8y$ ?

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Answer 1 · 2018-05-24T03:05:22

$4 {(a + b)}^{2}$ $4(a + b)^2$

$4 z + 8 y$ $4z+8y$

$a^{2} + b^{2} = z and a b = y$ $a^2 + b^2 = z and ab =y$

do some substitution:

$4 (a^{2} + b^{2}) + 8 (a b)$ $4(a^2 + b^2 )+8( ab)$

$4 a^{2} + 4 b^{2} + 8 a b$ $4a^2 + 4b^2 +8ab$

$4 (a^{2} + 2 a b + b^{2})$ $4(a^2 +2ab+ b^2)$

$4 (a + b) (a + b)$ $4(a + b)(a + b)$

$4 {(a + b)}^{2}$ $4(a + b)^2$

Answer 2 · 2018-05-24T03:15:11

$4 {(a + b)}^{2}$ $4(a+b)^2$

This is a substitution and simplification problem. Begin by taking the information we're given substitute z and y.

If $z = a^{2} + b^{2}$ $z=a^2+b^2$ and $y = a b$ $y=ab$ , then $4 z + 8 y = 4 (a^{2} + b^{2}) + 8 (a b)$ $4z+8y=4(a^2+b^2)+8(ab)$

From here, we distribute 4 into $(a^{2} + b^{2})$ $(a^2+b^2)$ and 8 into $(a b)$ $(ab)$ .

After the distribution we have $4 a^{2} + 4 b^{2} + 8 a b$ $4a^2+4b^2+8ab$

Rearrange our terms $4 a^{2} + 8 a b + 4 b^{2}$ $4a^2+8ab+4b^2$

From here, we need to recognize that this is the expanded form of a binomial raised to a second power.

Factor out a 4 from all 3 terms, and condense the expansion back into the binomial form.

$4 (a^{2} + 2 a b + b^{2}) \to 4 {(a + b)}^{2}$ $4(a^2+2ab+b^2) -> 4(a+b)^2$

If a2+b2=zandab=ya^2 + b^2 = z and ab =y, what will be the equivalent of 4z+8y4z+8y?