Find the roots of the quadratic polynomials:6x^2+5x-1 , 4x^2+4x-4

Cheyanne Leigh 2021-09-07 Answered
Find the roots of the quadratic polynomials:
\(\displaystyle{6}{x}^{{2}}+{5}{x}−{1}\)
\(\displaystyle{4}{x}^{{2}}+{4}{x}−{4}\)

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Expert Answer

Laaibah Pitt
Answered 2021-09-08 Author has 26433 answers
Given:
\(\displaystyle{6}{x}^{{2}}+{5}{x}-{1}\)
The roots of \(\displaystyle{a}{x}^{{2}}+{b}{x}+{c}\) are
\(\displaystyle{x}={\frac{{-{b}\pm\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}}\)
The roots of
\(\displaystyle{6}{x}^{{2}}+{5}{x}-{1}\)
\(\displaystyle{x}={\frac{{-{5}\pm\sqrt{{{5}^{{2}}-{4}{\left({6}\right)}{\left(-{1}\right)}}}}}{{{2}{\left({6}\right)}}}}\)
\(\displaystyle{x}={\frac{{-{5}\pm\sqrt{{{25}+{24}}}}}{{{12}}}}\)
\(\displaystyle{x}={\frac{{-{5}\pm\sqrt{{{49}}}}}{{{12}}}}\)
\(\displaystyle{x}={\frac{{-{5}\pm{7}}}{{{12}}}}\)
\(\displaystyle{x}=-{\frac{{{12}}}{{{12}}}},{\frac{{{2}}}{{{12}}}}\)
\(\displaystyle{x}=-{1},{\frac{{{1}}}{{{6}}}}\)
The roots of \(\displaystyle{6}{x}^{{2}}+{5}{x}-{1}\) are x=-1,\(\displaystyle{\frac{{{1}}}{{{6}}}}\)
The roots of
\(\displaystyle{4}{x}^{{2}}+{4}{x}-{4}\)
\(\displaystyle{x}={\frac{{-{4}\pm\sqrt{{{4}^{{2}}-{4}{\left({4}\right)}{\left(-{4}\right)}}}}}{{{2}{\left({4}\right)}}}}\)
\(\displaystyle{x}={\frac{{-{4}\pm\sqrt{{{16}+{64}}}}}{{{8}}}}\)
\(\displaystyle{x}={\frac{{-{4}+\sqrt{{{80}}}}}{{{8}}}}\)
\(\displaystyle{x}={\frac{{-{4}\pm{4}\sqrt{{{5}}}}}{{{8}}}}\)
\(\displaystyle{x}=-{\frac{{{1}}}{{{2}}}}+{\frac{{\sqrt{{{5}}}}}{{{2}}}},-{\frac{{{1}}}{{{2}}}}-{\frac{{\sqrt{{{5}}}}}{{{2}}}}\)
The roots of \(\displaystyle{4}{x}^{{2}}+{4}{x}-{4}\) are \(\displaystyle{x}=-{\frac{{{1}}}{{{2}}}}+{\frac{{\sqrt{{{5}}}}}{{{2}}}},-{\frac{{{1}}}{{{2}}}}-{\frac{{\sqrt{{{5}}}}}{{{2}}}}\)
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