Q17. - Algebra 1

Question

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

$2x^2+x-15=0$

Step-by-Step Solution

Answer

The solutions of the quadratic equations are $\mathbf{-}\mathbf{3}$ and 2.5.

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 equation and then making groups and equating them to zero.

Step 2. Substitute the values.

It is given that $2x^2+x-15=0$. Using quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, substitute 2 for a, 1 for b and $-15$ for c. 

Put the values in the formula and get $x=\frac{-1\pm \sqrt{{\left(1\right)}^2-4\times \left(2\right)\times \left(-15\right)}}{2\left(2\right)}$.

Step 3. Write the solutions.

Simplify the expression for x.

$$\begin{gathered} x=\frac{-1\pm \sqrt{{\left(1\right)}^2-4\times \left(2\right)\times \left(-15\right)}}{2\left(2\right)} \\ =\frac{-1\pm \sqrt{121}}{4} \\ =\frac{-1\pm 11}{4} \end{gathered}$$ 

 

Therefore, either

$$\begin{gathered} x=\frac{-1+11}{4} \\ =\frac{5}{2} \\ =2.5 \end{gathered}$$ 

 

or, 

 

$$\begin{gathered} x=\frac{-1-11}{4} \\ =-3 \end{gathered}$$ 

that is., either $x=-3$ or $x=2.5$.