Both sets of answers are correct. I only wonder if in the second one, you should have used the sq. root property. If you did it that way you would get the exact same answer. I just don't want you to get marked down for not following directions.
here's how to do it using sq. roots:
2x^2-35=15
2x^2=50
x^2=25 <-- sq. root both sides
x=+/- 5
Hello all - I would like to check my work on 2 problems.
Solve by factoring :
x^2-13x= -12
x^2-13x+12= -12 + 12
x^2-13x+12= 0
(x-1)(x-12)= 0
x = 1 , 12
Solve using the square root property:
2x^2-35=15
2x^2-35-15=0
2x^2-50=0
2(x-5)(x+5)=0
x = 5 , -5 Not sure I did this one right.
3 answers
Thankyou. I will use the square root way.
Wierd how I got the same answer.
Wierd how I got the same answer.
You're welcome.
If you think about it, it makes since why you got the same answer. The way you did it was right, just not the method they were asking for. There is never just one way to do things in math. :-)
If you think about it, it makes since why you got the same answer. The way you did it was right, just not the method they were asking for. There is never just one way to do things in math. :-)
0 answers