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

**looi76** · March 29th 2008, 11:35 AM

Question:
Write $x^2 + 10x + 38$ in the form $(x+b)^2+c$ where the values of b and c are to be found.
(a) State the minimum value of $x^2 + 10x + 38$ and the value of $x$ for which this occurs.
(b) Determine the values of $x$ for which $x^2 + 10x + 38 \geq 22$ .

Attempt:

$x^2 + 10x + 38$

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

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

What to do next?

$= (x + 5)^2 + 13$

**Moo** · March 29th 2008, 11:43 AM

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

**o_O** · March 29th 2008, 11:48 AM

For a parabola in the form of $y = a(x - b)^{2} + c$ (note the negative sign in front of b), the vertex is given by (b, c).

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

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

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