anonymous
  • anonymous
Solve the equation algebraically 5rad(x+4) =absolute value (x-3)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[5\sqrt{x+4}=\left| x-3 \right|\]
jdoe0001
  • jdoe0001
\(\large { 5\sqrt{x+4}=| x-3 |\\\quad\\ \implies \begin{cases} +(5\sqrt{x+4})=x-3\implies 5\sqrt{x+4}=x-3\\ 5\sqrt{x+4}+3=x \\\hline\\ -(5\sqrt{x+4})=x-3\implies -5\sqrt{x+4}=x-3\\ -5\sqrt{x+4}+3=x \end{cases} }\) 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
anonymous
  • anonymous
ok i got that but I still dont know how to solve this

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jdoe0001
  • jdoe0001
ohh shoot... aehmm yeah... ok... we do have an "x" on the left side... I see
jdoe0001
  • jdoe0001
have you ... covered the factoring quadratics?
anonymous
  • anonymous
idk maybe
jdoe0001
  • jdoe0001
hmmm meaning, have you factored something like \(x^2+3x-4\) yet?
anonymous
  • anonymous
yeah
jdoe0001
  • jdoe0001
ok.. have you covered the quadratic formula yet?
anonymous
  • anonymous
yass
jdoe0001
  • jdoe0001
k
jdoe0001
  • jdoe0001
let us pick the 1st case, one sec
jdoe0001
  • jdoe0001
\(\bf 5\sqrt{x+4}=x-3\impliedby \textit{let us square both sides} \\ \quad \\ (5\sqrt{x+4})^2=(x-3)^2\implies (5)^2(\sqrt{x+4})^2=x^2-6x+9 \\ \quad \\ 25(x+4)=x^2-6x+9\implies 25x+100=x^2-6x+9 \\ \quad \\ 0=x^2-31x-91\) 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
jdoe0001
  • jdoe0001
\(\bf 0=x^2-31x-91\qquad \qquad \textit{quadratic formula} \\ \quad \\ 0={\color{blue}{ 1}}x^2{\color{red}{ -31}}x{\color{green}{ -91}} \qquad \qquad x= \cfrac{ - {\color{red}{ b}} \pm \sqrt { {\color{red}{ b}}^2 -4{\color{blue}{ a}}{\color{green}{ c}}}}{2{\color{blue}{ a}}}\)
jdoe0001
  • jdoe0001
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
anonymous
  • anonymous
ok thank you sooooo much @jdoe0001
jdoe0001
  • jdoe0001
ywq
jdoe0001
  • jdoe0001
yw rather

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