Asked by Emory
Solve the quadratic equation.
13) x2 - x = 72
A) -8, 9
B) -8, -9
C) 1, 72
D) 8, 9
Can some one please check my answer and let me know if I did it the right way.
x2-x-72=0
(x-9)(x-8)=0
so the answer is D
8,9
13) x2 - x = 72
A) -8, 9
B) -8, -9
C) 1, 72
D) 8, 9
Can some one please check my answer and let me know if I did it the right way.
x2-x-72=0
(x-9)(x-8)=0
so the answer is D
8,9
Answers
Answered by
Reiny
no, your factors should have been
(x-9)(x+8) = 0
so x=9 or x=-8 which would be A
(x-9)(x+8) = 0
so x=9 or x=-8 which would be A
Answered by
Emory
Well you see that was how I did at first but I was not too sure about it so then I got it right the first time I always get confused with that any tips to remind me next time. Thanks
Answered by
Reiny
After you factor, you should multiply it back out to see if you end up with the original.
I knew immediately that your factors were wrong.
Both your factors ended with negatives for the constants, so your last term would have been positive, but it was -72
I knew immediately that your factors were wrong.
Both your factors ended with negatives for the constants, so your last term would have been positive, but it was -72
Answered by
Emory
Oh,I see thank you I have another one for you that I don't know how to start it and it goes like this 6y2 = 7y - 5?
Can you please help me just to start it off
Can you please help me just to start it off
Answered by
Reiny
put everything to one side
6y^2 - 7y + 5 = 0
I see a positive at the end, which tells me that the signs are the same, looking at the middle term they must be both negative.
so it must be (?x - 1)(?x - 5) = 0
only 4 possibilies are
(2x-1)(3x-5) = 0
(3x-1)(2x-5) = 0
(6x-1)(x-5) or
(x-1)(6x-5) = 0
unfortunately none of these give us a middle term of -7y
so your question does not factor.
6y^2 - 7y + 5 = 0
I see a positive at the end, which tells me that the signs are the same, looking at the middle term they must be both negative.
so it must be (?x - 1)(?x - 5) = 0
only 4 possibilies are
(2x-1)(3x-5) = 0
(3x-1)(2x-5) = 0
(6x-1)(x-5) or
(x-1)(6x-5) = 0
unfortunately none of these give us a middle term of -7y
so your question does not factor.
Answered by
Emory
So then the answer is 0 because like you mention there is no factor "correct"
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