Q41. - Algebra 1

Question

Solve the equation by using the Quadratic formula. Round to the nearest tenth if necessary.

$2f^2+7f-15=0$

Step-by-Step Solution

Answer

The solutions of the quadratic equations are $\frac{\mathbf{3}}{\mathbf{2}}$ and  $\mathbf{-}\mathbf{5}$. 

Step 1. Understand the concept used .

If $a{x}^2+bx+c=0$, with $a\ne 0$, then $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. Solving using factoring can be done by splitting the middle term of the equation and then making groups and equating them to zero.

Step 2. Substitute the values .

It is given that $2f^2+7f-15=0$. Using the quadratic formula $f=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, substitute 2 for $a, 7$ for $b$ and $-15$ for $c$.

Put the values in the formula and get $f=\frac{-7\pm \sqrt{{\left(7\right)}^2-4\times \left(2\right)\times \left(-15\right)}}{2\left(2\right)}$. 

Step 3. Write the solutions.

Simplify the expression for $f$.

$f=\frac{-7\pm \sqrt{{\left(7\right)}^2-4\times \left(2\right)\times \left(-15\right)}}{2\left(2\right)}$

$$\begin{gathered} =\frac{-7\pm \sqrt{49+120}}{4} \\ =\frac{-7\pm \sqrt{169}}{4} \\ =\frac{-7\pm 13}{4} \end{gathered}$$

Therefore, either

$$\begin{gathered} f=\frac{-7+13}{4} \\ =\frac{3}{2} \end{gathered}$$

or, 

$$\begin{gathered} f=\frac{-7-13}{4} \\ =-5 \end{gathered}$$

that is., either $f=\frac{3}{2}$ or $f=-5$.