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

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Answer 1 · 2018-05-24T03:05:22

#4(a + b)^2#

#4z+8y#

#a^2 + b^2 = z and ab =y#

do some substitution:

#4(a^2 + b^2 )+8( ab)#

#4a^2 + 4b^2 +8ab#

#4(a^2 +2ab+ b^2)#

#4(a + b)(a + b)#

#4(a + b)^2#

Answer 2 · 2018-05-24T03:15:11

#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# and #y=ab#, then #4z+8y=4(a^2+b^2)+8(ab)#

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

After the distribution we have #4a^2+4b^2+8ab#

Rearrange our terms #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+2ab+b^2) -> 4(a+b)^2#