Quadratic Equation: solve (y + 2)^2 - 3(y + 2) - 4 = 0
- Thread starter SaeLk
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Re: Quadratic Equation. Need Help Plz
Hello, SaeLk!
Let \(\displaystyle u \:=\:y\,+\,2\)
The equation becomes: \(\displaystyle \:u^2\,-\,3u\,-\,4\;=\;0\)
. . which factors: \(\displaystyle \
u\,+\,1)(u\,-\,4)\;=\;0\)
. . and has roots: \(\displaystyle \:\begin{array}{cc}u\,+\,1\:=\:0 & \;\;\Rightarrow\;\; & u \:=\:-1 \\ u\,-\,4\:=\:0 & \;\;\Rightarrow\;\; & u \:=\:4\end{array}\)
Since \(\displaystyle u \:=\:y\,+\,2\), we have: \(\displaystyle \:\begin{array}{cc}y\,+\,2\:=\:-1 & \;\;\Rightarrow\;\;& \fbox{y\:=\:-3} \\ y\,+\,2\:=\:4 & \;\;\Rightarrow\;\; & \fbox{y \:=\:2}\end{array}\)
Hello, SaeLk!
Let \(\displaystyle u \:=\:y\,+\,2\)
The equation becomes: \(\displaystyle \:u^2\,-\,3u\,-\,4\;=\;0\)
. . which factors: \(\displaystyle \
u\,+\,1)(u\,-\,4)\;=\;0\). . and has roots: \(\displaystyle \:\begin{array}{cc}u\,+\,1\:=\:0 & \;\;\Rightarrow\;\; & u \:=\:-1 \\ u\,-\,4\:=\:0 & \;\;\Rightarrow\;\; & u \:=\:4\end{array}\)
Since \(\displaystyle u \:=\:y\,+\,2\), we have: \(\displaystyle \:\begin{array}{cc}y\,+\,2\:=\:-1 & \;\;\Rightarrow\;\;& \fbox{y\:=\:-3} \\ y\,+\,2\:=\:4 & \;\;\Rightarrow\;\; & \fbox{y \:=\:2}\end{array}\)