# X^2+4x-5 can be written as (x+p)^2+q. Find P and Q?

## I know how to convert it into the final form but what i do not know is how to determine if the answer would be positive or negative

May 1, 2018

$p = 2$

$q = - 9$

#### Explanation:

Assuming your question is stated as below :

${x}^{2} + 4 x - 5$ can be written as ${\left(x + p\right)}^{2} + q$

$\implies {x}^{2} + 4 x - 5$ is equal to ${\left(x + 2\right)}^{2} - 9$

By comparison,

$\therefore p = 2 \mathmr{and} q = - 9$

May 1, 2018

$p = 2 \text{ and } q = - 9$

#### Explanation:

$\text{convert to vertex form using "color(blue)"completing the square}$

•color(white)(x)y=a(x-h)^2+klarrcolor(blue)"vertex form"

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

• " the coefficient of the "x^2 " term must be 1 which it is"

• " add/subtract "(1/2"coefficient of the x-term")^2" to"
${x}^{2} + 4 x$

$= {x}^{2} + 2 \left(2\right) x \textcolor{red}{+ 4} \textcolor{red}{- 4} - 5$

$= {\left(x + 2\right)}^{2} - 9 \leftarrow \textcolor{red}{\text{in vertex form}}$

$\Rightarrow - h = 2 \Rightarrow h = - 2 \text{ and } k = - 9$

$\text{compare with "(x+p)^2+qrArrp=2" and } q = - 9$
graph{(y-x^2-4x+5)((x+2)^2+(y+9)^2-0.04)=0 [-20, 20, -10, 10]}