Q42. - Algebra 1

Question

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

$2h^2+8h+3=3$

Step-by-Step Solution

Answer

The solutions of the quadratic equations are 0 and $-4$. 

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 $2h^2+8h+3=3$. The equation can be rewritten as $2h^2+8h=0$. Using the quadratic formula $h=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, substitute 2 for $a, 8$ for $b$ and 0 for $c$.

Put the values in the formula and get $h=\frac{-8\pm \sqrt{{\left(8\right)}^2-4\times \left(2\right)\times \left(0\right)}}{2\left(2\right)}$. 

Step 3. Write the solutions.

Simplify the expression for $h$.

 $h=\frac{-8\pm \sqrt{{\left(8\right)}^2-4\times \left(2\right)\times \left(0\right)}}{2\left(2\right)}$

   $$\begin{gathered} =\frac{-8\pm \sqrt{64}}{4} \\ =\frac{-8\pm 8}{4} \end{gathered}$$

Therefore, either

 $$\begin{gathered} h=\frac{-8+8}{4} \\ =0 \end{gathered}$$

or,

$$\begin{gathered} h=\frac{-8-8}{4} \\ =-4 \end{gathered}$$

that is., either $h=0$ or $h=-4$.