How do you write 4(x+1)2+1 in the form (ax+b)2+c?

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Killian Curry · Accepted Answer

You can also solve the equation(ax+b)2+c≡4x2+8x+15 (the ≡ sign means &quot;true for all values of x&quot;-the coefficients on both sides of the equation have to be the same). If we expand the left-hand side, we get(ax+b)2+c≡a2x2+2abx+b2+c≡4x2+8x+5,.This means a2=4 and 2ab=8 and b2+c=15. If a2=4 then a=2 or a=−2. Suppose that a=2. Then, 4b=8, and so b=2. Finally, 4+c=5, and so c=1. This gives us the solution(2x+2)2+1≡4x2+8x+5,.If a=−2, then −4b=8, and so b=−2. Finally, 4+c=5, and so c=1. This gives us the solution(−2x−2)2+1≡4x2+8x+5,.However, the first solution is arguably simpler because it doesn&#039;t have any minus signs in it.

aimadorsozf · Accepted Answer

If one has already found a(x+b)2+c form, then provided a&amp;gt;0 one can start by re-writing a as [a]2 and use that u2v2=(uv)2 to get to(a x+a b)2+c.There is no need for the &quot;unsightly&quot; square root symbol. So (2x+2)2+1.

How do you write 4(x+1)^{2}+1 in the form (ax+b)^{2}+c?

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