Find the roots of the quadratic polynomials: a) f(x)=4x^2-3x-1. b) f(x)=x^2-2x-1

Ramsey 2021-09-19 Answered
Find the roots of the quadratic polynomials:
a) \(\displaystyle{f{{\left({x}\right)}}}={4}{x}^{{2}}-{3}{x}-{1}\)
b) \(\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}-{2}{x}-{1}\)

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Expert Answer

Nola Robson
Answered 2021-09-20 Author has 6268 answers
We set them equal to 0 and solve for x
we first factor the ledt side
\(\displaystyle{4}{x}^{{2}}-{3}{x}-{1}={0}\)
\(\displaystyle{4}{x}^{{2}}-{4}{x}+{x}-{1}={0}\)
\(\displaystyle{4}{x}{\left({x}-{1}\right)}+{1}{\left({x}-{1}\right)}={0}\)
\(\displaystyle{\left({x}-{1}\right)}{\left({4}{x}+{1}\right)}={0}\)
then we set rach factor equal to 0 and find x
\(\displaystyle{x}-{1}={0}\)
\(\displaystyle{x}={1}\)
\(\displaystyle{4}{x}+{1}={0}\)
\(\displaystyle{4}{x}=-{1}\)
\(\displaystyle{x}=-{\frac{{{1}}}{{{4}}}}\)
b) This expression can not be factored so we use the quadratic formula
\(\displaystyle{x}^{{2}}-{2}{x}-{1}={0}\)
\(\displaystyle{x}={\frac{{-{\left(-{2}\right)}\pm\sqrt{{{\left(-{2}\right)}^{{2}}-{4}{\left({1}\right)}{\left(-{1}\right)}}}}}{{{2}{\left({1}\right)}}}}\)
\(\displaystyle{x}={\frac{{{2}\pm\sqrt{{{8}}}}}{{{2}}}}\)
\(\displaystyle{x}={\frac{{{2}\pm{2}\sqrt{{{2}}}}}{{{2}}}}\)
\(\displaystyle{x}={1}\pm\sqrt{{{2}}}\)
Answer: \(\displaystyle{x}={1}\pm\sqrt{{{2}}}\)
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