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\((2x+5)(3x-4)=0\)

The solutions are \(-\frac{5}{2}\) and \(\frac{4}{3}\text{.}\)

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Use the zero-product property to solve each quadratic equation.

\((2x+5)(3x-4)=0\)

Solution

The solutions are \(-\frac{5}{2}\) and \(\frac{4}{3}\text{.}\)

\(4x(x+6)=0\)

Solution

The solutions are \(0\) and \(-6\)

\(x^2-10x-24=0\)

Solution

The solutions are \(12\) and \(-2\text{.}\)

\(x(x+7)=30\)

Solution

The solutions are \(-10\) and \(3\text{.}\)

\(12w^2=-9w\)

Solution

The solutions are \(0\) and \(-\frac{3}{4}\text{.}\)

\(25y^2+30y+9=0\)

Solution

The only solution is \(-\frac{3}{5}\text{.}\)

Use the square root method to solve each of the following quadratic equations. Make sure that all solutions are completely simplified. Note that for some of the equations you will need to first complete the square.

\((5t+7)^2=49\)

Solution

The solutions are \(0\) and \(-\frac{14}{5}\text{.}\)

\((2x-3)^2=12\)

Solution

The solutions are \(\frac{3-2\sqrt{3}}{2}\) and \(\frac{3+2\sqrt{3}}{2}\text{.}\)

\(x^2+8x-20=0\)

Solution

The solutions are \(-10\) and \(2\text{.}\)

\(y^2-20y+40=0\)

Solution

The solutions are \(10-2\sqrt{15}\) and \(10+2\sqrt{15}\text{.}\)

Use the quadratic formula to solve each of the following equations over the real numbers.

\(2x^2+8x-5=0\)

Solution

The solutions are \(\frac{-4-\sqrt{26}}{2}\) and \(\frac{-4+\sqrt{26}}{2}\text{.}\)

\(t^2-15t=100\)

Solution

The solutions are \(-5\) and \(20\text{.}\)

\(-x^2+4x+3=0\)

Solution

The solutions are \(2-\sqrt{7}\) and \(2+\sqrt{7}\text{.}\)

\(x(3x-5)=-6\)

Solution

The equation has no real number solutions,

Use the square root method to solve each of the following equations over the complex numbers. Make sure that all solutions have been completely simplified. Note that for one of the equations you will need to first complete the square.

Use the quadratic formula to solve each of the following equations over the complex numbers. Make sure that all solutions have been completely simplified.

For each equation, use the value of the discriminant to determine the nature of the solution. That's all - do not find the actual solutions.

\(12x^2+4x+3=0\)

Solution

\begin{equation*}
b^2-4ac=-128
\end{equation*}

The equation has two complex number solutions with non-zero imaginary coefficients.

\(-x^2+4x+4=0\)

Solution

\begin{equation*}
b^2-4ac=32
\end{equation*}

The equation has two real number solutions.

\(-49+14x-x^2=0\)

Solution

\begin{equation*}
b^2-4ac=0
\end{equation*}

The equation has one real number solution.