Step 1

Given : The quadratic equation \(\displaystyle{x}^{{{2}}}-{3}{x}-{54}={0}\)

we have to tell using the discriminant to determine whether the equation has two nonreal complex solutions, one real solution with a multiplicity of two, or two real solutions also we have to solve the given equation.

Step 2

as we know that the discriminant is given as \(\displaystyle{b}^{{{2}}}-{4}{a}{c}\)

here \(\displaystyle{x}^{{{2}}}-{3}{x}-{54}={0}\)

a=1, b=-3, c=-54

\(\displaystyle{D}={\left(-{3}\right)}^{{{2}}}-{4}{\left({1}\right)}{\left(-{54}\right)}\)

\(\displaystyle{D}={\left(-{3}\right)}^{{{2}}}-{4}{\left({1}\right)}{\left(-{54}\right)}\)

D=9+216

D=225

since D is a perfect square thereforeThe roots exists are real and distinct.

\(\displaystyle{x}={\frac{{-{b}\pm\sqrt{{{D}}}}}{{{2}{a}}}}\)

\(\displaystyle{x}={\frac{{-{\left(-{3}\right)}\pm\sqrt{{{225}}}}}{{{2}}}}\)

\(\displaystyle{x}={\frac{{{3}\pm{15}}}{{{2}}}}\)

\(\displaystyle{x}={\frac{{{3}+{15}}}{{{2}}}},{\frac{{{3}-{15}}}{{{2}}}}\)

\(\displaystyle{x}={\frac{{{18}}}{{{2}}}},-{\frac{{{12}}}{{{2}}}}\)

x=9, -6

Given : The quadratic equation \(\displaystyle{x}^{{{2}}}-{3}{x}-{54}={0}\)

we have to tell using the discriminant to determine whether the equation has two nonreal complex solutions, one real solution with a multiplicity of two, or two real solutions also we have to solve the given equation.

Step 2

as we know that the discriminant is given as \(\displaystyle{b}^{{{2}}}-{4}{a}{c}\)

here \(\displaystyle{x}^{{{2}}}-{3}{x}-{54}={0}\)

a=1, b=-3, c=-54

\(\displaystyle{D}={\left(-{3}\right)}^{{{2}}}-{4}{\left({1}\right)}{\left(-{54}\right)}\)

\(\displaystyle{D}={\left(-{3}\right)}^{{{2}}}-{4}{\left({1}\right)}{\left(-{54}\right)}\)

D=9+216

D=225

since D is a perfect square thereforeThe roots exists are real and distinct.

\(\displaystyle{x}={\frac{{-{b}\pm\sqrt{{{D}}}}}{{{2}{a}}}}\)

\(\displaystyle{x}={\frac{{-{\left(-{3}\right)}\pm\sqrt{{{225}}}}}{{{2}}}}\)

\(\displaystyle{x}={\frac{{{3}\pm{15}}}{{{2}}}}\)

\(\displaystyle{x}={\frac{{{3}+{15}}}{{{2}}}},{\frac{{{3}-{15}}}{{{2}}}}\)

\(\displaystyle{x}={\frac{{{18}}}{{{2}}}},-{\frac{{{12}}}{{{2}}}}\)

x=9, -6