# [SOLVED] Write x^2 + 10x + 38 in the form (x+b)^2+c...

• Mar 29th 2008, 11:35 AM
looi76
[SOLVED] Write x^2 + 10x + 38 in the form (x+b)^2+c...
Question:
Write $\displaystyle x^2 + 10x + 38$ in the form $\displaystyle (x+b)^2+c$ where the values of b and c are to be found.
(a) State the minimum value of $\displaystyle x^2 + 10x + 38$ and the value of $\displaystyle x$ for which this occurs.
(b) Determine the values of $\displaystyle x$ for which $\displaystyle x^2 + 10x + 38 \geq 22$.

Attempt:

$\displaystyle x^2 + 10x + 38$

$\displaystyle = (x + \frac{10}{2})^2 - (\frac{10}{2})^2 + 38$

$\displaystyle = (x + 5)^2 - 25 + 38$

What to do next?

$\displaystyle = (x + 5)^2 + 13$
• Mar 29th 2008, 11:43 AM
Moo
Hello,

The beginning is correct.

To continue, you can remember that something squared is always > 0.

So (x+5)²+13 has a minimum when (x+5)² = 0, because 0 is the minimal value for a square.

For b), (x+5)²+13>22 <=> (x+5)²>9=3²

-> |x+5|>3
• Mar 29th 2008, 11:48 AM
o_O
For a parabola in the form of $\displaystyle y = a(x - b)^{2} + c$ (note the negative sign in front of b), the vertex is given by (b, c).

$\displaystyle x^{2} + 10x + 38 \geq 22$
$\displaystyle x^{2} + 10x + 16 \geq 0$

Now solve the inequality by finding the roots and finding the regions in which the inequality holds.