## anonymous one year ago Solve the equation algebraically 5rad(x+4) =absolute value (x-3)

1. anonymous

$5\sqrt{x+4}=\left| x-3 \right|$

2. anonymous

$$\large { 5\sqrt{x+4}=| x-3 |\\\quad\\ \implies \begin{cases} +(5\sqrt{x+4})=x-3\implies 5\sqrt{x+4}=x-3\\ 5\sqrt{x+4}+3=x \\\hline\\ -(5\sqrt{x+4})=x-3\implies -5\sqrt{x+4}=x-3\\ -5\sqrt{x+4}+3=x \end{cases} }$$ two cases, for an absolute value expression, thus two values for "x" keep in mind you always have a + and - case with absolute value expressions

3. anonymous

ok i got that but I still dont know how to solve this

4. anonymous

ohh shoot... aehmm yeah... ok... we do have an "x" on the left side... I see

5. anonymous

have you ... covered the factoring quadratics?

6. anonymous

idk maybe

7. anonymous

hmmm meaning, have you factored something like $$x^2+3x-4$$ yet?

8. anonymous

yeah

9. anonymous

ok.. have you covered the quadratic formula yet?

10. anonymous

yass

11. anonymous

k

12. anonymous

let us pick the 1st case, one sec

13. anonymous

$$\bf 5\sqrt{x+4}=x-3\impliedby \textit{let us square both sides} \\ \quad \\ (5\sqrt{x+4})^2=(x-3)^2\implies (5)^2(\sqrt{x+4})^2=x^2-6x+9 \\ \quad \\ 25(x+4)=x^2-6x+9\implies 25x+100=x^2-6x+9 \\ \quad \\ 0=x^2-31x-91$$ now, we end up with a quadratic equation 91 is a prime, pretty sure of that so it won't factor nicely, so we could get the middle term of -6 so you'd need to use the quadratic formula for that

14. anonymous

$$\bf 0=x^2-31x-91\qquad \qquad \textit{quadratic formula} \\ \quad \\ 0={\color{blue}{ 1}}x^2{\color{red}{ -31}}x{\color{green}{ -91}} \qquad \qquad x= \cfrac{ - {\color{red}{ b}} \pm \sqrt { {\color{red}{ b}}^2 -4{\color{blue}{ a}}{\color{green}{ c}}}}{2{\color{blue}{ a}}}$$

15. anonymous

now let use pick the 2nd case $$\bf -5\sqrt{x+4}=x-3\impliedby \textit{let us square both sides} \\ \quad \\ (-5\sqrt{x+4})^2=(x-3)^2\implies (-5)^2(\sqrt{x+4})^2=x^2-6x+9 \\ \quad \\ 25(x+4)=x^2-6x+9\implies 25x+100=x^2-6x+9 \\ \quad \\ 0=x^2-31x-91$$ notice, the -5, turns to a positive 25 and we end up with the same equation as the 1st case so, using the quadratic formula for the 1st one, solves also the 2nd one

16. anonymous

ok thank you sooooo much @jdoe0001

17. anonymous

ywq

18. anonymous

yw rather