in this problem, it is very useful to use a substitution in this case the one that seems to work is set u=x^2 if we do that, you will see that it turns this into a polynomial of degree 2 in which you can factor by trial and error or apply quadratic formula if you do not see the factors lets get started :
let u = x^2
we have u^2-5u+6=0
factor
(u - 3) (u- 2) = 0 so solutions for U are u = 3 and U = 2
since we let u = x ^2 we now must solve for x the following two equations.
x^2 = 3 and x ^2 = 2
x = sqrt (3) and x = sqrt (2)
a common question that is asked is " how do i know that I have substituted correctly ?" notice how we are able to place the whole equation in terms of U. If you cannot replace all the variables with the substitution your solution would have not been correct.
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Verified answer
y=x^2
y*2 - 5y + 6=0
y1=3 => x1=sqrt3 x2= - sqrt3
y2=2 => x3=sqrt2 x4= - sqrt2
in this problem, it is very useful to use a substitution in this case the one that seems to work is set u=x^2 if we do that, you will see that it turns this into a polynomial of degree 2 in which you can factor by trial and error or apply quadratic formula if you do not see the factors lets get started :
let u = x^2
we have u^2-5u+6=0
factor
(u - 3) (u- 2) = 0 so solutions for U are u = 3 and U = 2
since we let u = x ^2 we now must solve for x the following two equations.
x^2 = 3 and x ^2 = 2
x = sqrt (3) and x = sqrt (2)
a common question that is asked is " how do i know that I have substituted correctly ?" notice how we are able to place the whole equation in terms of U. If you cannot replace all the variables with the substitution your solution would have not been correct.
Let y=x^2
The equation may be written as y^2-5y+6=0
y^2 -3y-2y+6=0
y(y-3)-2(y-3)=0
(y-3)(y-2)=0
y=3
y=2
x^2 = 3 means x= +/- sqrt(3)
x^2 = 2 means x= +/- sqrt(2)
The four solutions are -sqrt(2),sqrt(2),-sqrt(3), and sqrt(3)