SOLUTION: I have to solve this: √x-1=x-3
This is what I have done so far. Am I on the right track? What do I do now?
"The radical symbol is only over the x-1"
√x-1=x-3
(W
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-> SOLUTION: I have to solve this: √x-1=x-3
This is what I have done so far. Am I on the right track? What do I do now?
"The radical symbol is only over the x-1"
√x-1=x-3
(W
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Question 82585: I have to solve this: √x-1=x-3
This is what I have done so far. Am I on the right track? What do I do now?
"The radical symbol is only over the x-1"
√x-1=x-3
(√x-1)^2=(x-3)^2
x-1 = (x-3)(x-3)
x-1 = x^2-6x+9
x-1 = x^2-6x+9
0=x^2-7x+10
You can put this solution on YOUR website!
GOOD WORK!!!!!!!!YOU ARE ALMOST FINISHED
You now have a quadratic in standard form where:
A=1
B=-7
C=10
We can solve it using the quadratic formula but, by inspection, we can see that this quadratic can be factored. The factors are:
(x-2)(x-5)=0
x-2=0
x=2
and
x-5=0
x=5
(Note: If the A coefficient of a quadratic equals 1 then the B coefficient will be the sum of the factors of the C coefficient, if the quadratic is readily factorable. In this case, the factors of the C coefficient that works are -5 and -2.)
Hope this helps---ptaylor
You can put this solution on YOUR website! Good job!! I agree with your answer. You correctly initially squared both sides of the equation in order to get rid of the radical; used FOIL on the left side; and solved for the x-term.
.
If your instructor/book wanted you to solve for the x=term, just continue the process:
0=x^2-7x+10
(x-5)(x-2)=0
x-5=0
x=5
or x-2=0
x=2
.
Try plugging each x-value back into the original equation to make sure each solution is a true solution. One of the values of x could possibly be an erroneous solution:
. =
Let x=5 = =[the square root of 4 is either a plus or minus 2]
2=2 [so, x=5 checks out]
. =
Let x=2 = [the square root of 1 is either a plus or minus (1). In this case, the (-1) works]
-1 = -1 [ture, so x=2 is also a solution]
