Question

part 5 of 6
rounding to the nearest tenth,
$x = \frac{-8 + \sqrt{224}}{2} \approx \square$
$x = \frac{-8 - \sqrt{224}}{2} \approx \square$

Explanation:

Step1: Calculate the square root of 224

First, we find the value of $\sqrt{224}$. We know that $14^2 = 196$ and $15^2 = 225$, so $\sqrt{224}\approx14.9666$.

Step2: Calculate the first solution

For $x=\frac{-8 + \sqrt{224}}{2}$, substitute $\sqrt{224}\approx14.9666$ into the formula:
\[

$$\begin{align*} x&=\frac{-8 + 14.9666}{2}\\ &=\frac{6.9666}{2}\\ &= 3.4833\\ &\approx3.5 \quad (\text{rounding to the nearest tenth}) \end{align*}$$

\]

Step3: Calculate the second solution

For $x=\frac{-8 - \sqrt{224}}{2}$, substitute $\sqrt{224}\approx14.9666$ into the formula:
\[

$$\begin{align*} x&=\frac{-8 - 14.9666}{2}\\ &=\frac{-22.9666}{2}\\ &=- 11.4833\\ &\approx - 11.5 \quad (\text{rounding to the nearest tenth}) \end{align*}$$

\]

Answer:

For $x=\frac{-8 + \sqrt{224}}{2}\approx\boldsymbol{3.5}$; for $x=\frac{-8 - \sqrt{224}}{2}\approx\boldsymbol{-11.5}$