How do you solve $p^{2} + 3 p - 9 = 0$ $p^2 + 3p - 9 = 0$ by completing the square?

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How do you solve $p^{2} + 3 p - 9 = 0$ $p^2 + 3p - 9 = 0$ by completing the square?

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Answer 1 · 2017-05-28T07:58:13

$p = - \frac{3}{2} - \frac{\sqrt{45}}{2}$ $p=-3/2-sqrt(45)/2$ and $p = - \frac{3}{2} + \frac{\sqrt{45}}{2}$ $p=-3/2+sqrt(45)/2$

Explanation:

Step 1. Add and subtract the perfect square term.

The perfect square term starts with the value next to the variable $p$ $p$ , in this case $3$ $3$ . First, you cut it in half:

$3 \Rightarrow \frac{3}{2}$ $3 => 3/2$

Then you square the result

$\frac{3}{2} \Rightarrow {(\frac{3}{2})}^{2} = \frac{9}{4}$ $3/2 => (3/2)^2=9/4$

Then add and subtract this term in the original expression.

$p^{2} + 3 p + \frac{9}{4} - \frac{9}{4} - 9 = 0$ $p^2+3p+9/4-9/4-9=0$

Step 2. Factor the perfect square.

The terms in $blue$ $color(blue)("blue")$ are the perfect square.

$p^{2} + 3 p + \frac{9}{4} - \frac{9}{4} - 9 = 0$ $color(blue)(p^2+3p+9/4)-9/4-9=0$

${(p + \frac{3}{2})}^{2} - \frac{9}{4} - 9 = 0$ $color(blue)((p+3/2)^2)-9/4-9=0$

Step 3. Simplify the remaining terms in $red$ $color(red)("red")$ .

${(p + \frac{3}{2})}^{2} - \frac{9}{4} - 9 = 0$ $(p+3/2)^2color(red)(-9/4-9)=0$

${(p + \frac{3}{2})}^{2} - \frac{45}{4} = 0$ $(p+3/2)^2color(red)(-45/4)=0$

Step 4. Solve for $p$ $p$ .

${(p + \frac{3}{2})}^{2} - \frac{45}{4} = 0$ $(p+3/2)^2-45/4=0$

Add $45 / 4$ $45//4$ to both sides

${(p + \frac{3}{2})}^{2} = \frac{45}{4}$ $(p+3/2)^2=45/4$

Take $\pm$ $+-$ the square root of both sides.

$p + \frac{3}{2} = \pm \sqrt{\frac{45}{4}}$ $p+3/2=+-sqrt(45/4)$

Subtract $3 / 2$ $3//2$ from both sides.

$p = - \frac{3}{2} \pm \frac{\sqrt{45}}{2}$ $p=-3/2+-sqrt(45)/2$

$p = - \frac{3}{2} - \frac{\sqrt{45}}{2}$ $p=-3/2-sqrt(45)/2$ and $p = - \frac{3}{2} + \frac{\sqrt{45}}{2}$ $p=-3/2+sqrt(45)/2$

Answer 2 · 2017-05-28T07:59:06

$p = - \frac{3}{2} \pm \frac{3 \sqrt{5}}{2}$ $p=-3/2+-[3sqrt(5)]/2$

Explanation:

$p^{2} + 3 p - 9 = 0$ $p^2+ 3p-9=0$ => add 9 to both sides:
$p^{2} + 3 p = 9$ $p^2+3p=9$ => using; ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2=a^2+ 2ab+b^2$
$p^{2} + 3 p + {(\frac{3}{2})}^{2} = 9 + \frac{9}{4}$ $p^2+3p+(3/2)^2= 9 + 9/4$
${(p + \frac{3}{2})}^{2} = \frac{45}{4}$ $(p+3/2)^2=45/4$ => take square root of both sides:
$p + \frac{3}{2} = \pm \frac{\sqrt{45}}{2}$ $p+3/2=+-sqrt(45)/2$ => simplify:
$p = - \frac{3}{2} \pm \frac{3 \sqrt{5}}{2}$ $p=-3/2+-[3sqrt(5)]/2$

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How do you solve $p^{2} + 3 p - 9 = 0$ $p^2 + 3p - 9 = 0$ by completing the square?

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