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Mathematics 7 Online
OpenStudy (anonymous):

need a solutions √(y+10)=y-2

OpenStudy (anonymous):

(y+10)=(y-2)^2 y+10=y^2-4y+4 y^2-5y-6=0 can you solve this quadratic equation?

OpenStudy (anonymous):

nop

OpenStudy (anonymous):

use quadratic formula: \[y=(5\pm \sqrt{5^{2}-4*(-6)})/2*1=(5\pm7)/2\]

OpenStudy (anonymous):

y1=6 and y2=-1 this are the roots of quadratic equation/ solution(s) now you need to check if both solutions are good - put the values in original equation: if y=6, then: \[\sqrt{6+10}=6-2\] 4=4 is a true statement the root is good. if y=-1, then: \[\sqrt{-1+10}=-1-2\] 3=-3 is not a true statement, so y=-1 is not a solution. Answer: y=6 (only one solution)

OpenStudy (anonymous):

thanks alot

OpenStudy (anonymous):

sure:)

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