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

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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Nola Robson
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: $$\displaystyle{x}={1}\pm\sqrt{{{2}}}$$