Find the roots of the quadratic polynomials:f(x)=x^2-2x-1

defazajx 2021-09-18 Answered
Find the roots of the quadratic polynomials:
\(\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}-{2}{x}-{1}\)

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Expert Answer

wornoutwomanC
Answered 2021-09-19 Author has 8427 answers

We know that, for an standard equation \(\displaystyle{f{{\left({x}\right)}}}={a}{x}^{{2}}+{b}{x}+{c}\), roots are calculated by using quadratic formula as \(\displaystyle{x}={\frac{{-{b}\pm\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}}\)
We have the given equation as
\(\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}-{2}{x}-{1}\)
On comparing the equation \(\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}-{2}{x}-{1}\) with standard equation \(\displaystyle{f{{\left({x}\right)}}}={a}{x}^{{2}}+{b}{x}+{c}\), we get the result as \(\displaystyle{a}={1},{b}=-{2},{\quad\text{and}\quad}{c}=-{1}\)
Therefore, by using equation, we get the zeros as
\(\displaystyle{x}={\frac{{-{\left(-{2}\right)}\pm\sqrt{{{\left(-{2}\right)}^{{2}}-{4}{\left({1}\right)}{\left(-{1}\right)}}}}}{{{\left({2}\right)}{\left({1}\right)}}}}\)
\(\displaystyle{x}={\frac{{{2}\pm\sqrt{{{4}+{4}}}}}{{{2}}}}\)
\(\displaystyle{x}={\frac{{{2}\pm{2}\sqrt{{{2}}}}}{{{2}}}}\)
\(\displaystyle{x}={1}\pm\sqrt{{{2}}}\)
Hence, zeros of the quadratic \(\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}-{2}{x}-{1}\) are \(\displaystyle{x}={1}+\sqrt{{{2}}}\)  and  \(\displaystyle{x}={1}-\sqrt{{{2}}}\)

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