Express x^{2}+8x in the form (x+a)2+b

Question
Quadratics
asked 2021-03-02
Express \(\displaystyle{x}^{{{2}}}+{8}{x}\) \ in the form \ \(\displaystyle{\left({x}+{a}\right)}{2}+{b}\)

Answers (1)

2021-03-03
To express \(\displaystyle{x}^{{{2}}}+{8}{x}\) in the form (x+a)2+b, we need to complete the square.
For an expression of the form \(\displaystyle{x}^{{{2}}}+{B}{x}\), we can complete the square by adding and subtracting \(\displaystyle{\frac{{{B}}}{{{2}}}}^{{{2}}}\). You need to add and subtract this number so the new expression will be equivalent to the original expression.
For x2+8x, B=8 so we need to add and subtract (82)2=42=16. This gives:
\(\displaystyle{x}^{{{2}}}+{8}{x}={x}^{{{2}}}+{8}{x}+{16}−{16}\)
Group the first three terms to form a perfect square trinomial:
\(\displaystyle={\left({x}^{{2}}+{8}{x}+{16}\right)}−{16}\)
A perfect square trinomial of the form \(\displaystyle{x}^{{{2}}}+{B}{x}+{\frac{{{B}}}{{{2}}}}^{{{2}}}\) factors to \(\displaystyle{\left({x}+{B}{2}\right)}{2}{\left({x}+{2}{B}\right)}^{{{2}}}\).
Since \(\displaystyle{\frac{{{B}}}{{{2}}}}={\frac{{{8}}}{{{2}}}}={4},{t}{h}{e}{n}\ {x}^{{{2}}}+{8}{x}+{16}\) factors to \(\displaystyle{\left({x}+{4}\right)}^{{{2}}}\). Therefore:
\(\displaystyle{\left({x}^{{{2}}}+{8}{x}+{16}\right)}−{16}={\left({x}+{4}\right)}^{{{2}}}−{16}\)
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