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anonymous
 one year ago
Solve the equation algebraically
5rad(x+4) =absolute value (x3)
anonymous
 one year ago
Solve the equation algebraically 5rad(x+4) =absolute value (x3)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[5\sqrt{x+4}=\left x3 \right\]

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1\(\large { 5\sqrt{x+4}= x3 \\\quad\\ \implies \begin{cases} +(5\sqrt{x+4})=x3\implies 5\sqrt{x+4}=x3\\ 5\sqrt{x+4}+3=x \\\hline\\ (5\sqrt{x+4})=x3\implies 5\sqrt{x+4}=x3\\ 5\sqrt{x+4}+3=x \end{cases} }\) 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok i got that but I still dont know how to solve this

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1ohh shoot... aehmm yeah... ok... we do have an "x" on the left side... I see

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1have you ... covered the factoring quadratics?

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1hmmm meaning, have you factored something like \(x^2+3x4\) yet?

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1ok.. have you covered the quadratic formula yet?

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1let us pick the 1st case, one sec

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1\(\bf 5\sqrt{x+4}=x3\impliedby \textit{let us square both sides} \\ \quad \\ (5\sqrt{x+4})^2=(x3)^2\implies (5)^2(\sqrt{x+4})^2=x^26x+9 \\ \quad \\ 25(x+4)=x^26x+9\implies 25x+100=x^26x+9 \\ \quad \\ 0=x^231x91\) 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

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1\(\bf 0=x^231x91\qquad \qquad \textit{quadratic formula} \\ \quad \\ 0={\color{blue}{ 1}}x^2{\color{red}{ 31}}x{\color{green}{ 91}} \qquad \qquad x= \cfrac{  {\color{red}{ b}} \pm \sqrt { {\color{red}{ b}}^2 4{\color{blue}{ a}}{\color{green}{ c}}}}{2{\color{blue}{ a}}}\)

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1now let use pick the 2nd case \(\bf 5\sqrt{x+4}=x3\impliedby \textit{let us square both sides} \\ \quad \\ (5\sqrt{x+4})^2=(x3)^2\implies (5)^2(\sqrt{x+4})^2=x^26x+9 \\ \quad \\ 25(x+4)=x^26x+9\implies 25x+100=x^26x+9 \\ \quad \\ 0=x^231x91\) 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok thank you sooooo much @jdoe0001
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