Asked by BelovedAngel
How would I solve this one? I just can't get this one.
6x^2+11x-35=0
6x^2+11x-35=0
Answers
Answered by
Reiny
Try to factor it as Steve told you in a previous post along the same type of question
What methods have you learned?
A common method these days seems to be "decomposition"
multiply the first and last coefficients:
6(-35) = -210
Now look for 2 factors which when multiplied will give you -210 and when added will give you +11
how about 21 and -10
so decompose the middle term of 11x into -10x + 21x
6x^2 - 10x + 21x - 35 = 0
2x(3x - 5) + 7(3x - 5) = 0
(3x-5)(2x + 7) = 0
x = 5/3 or x = -7/2
or
by formula:
x = (-11 ± √(121 - 4(6)(-35)) )/12
= (-11 ± √961)/12
= (-11 ± 31)/12
= 5/3 or -7/2
same as above
What methods have you learned?
A common method these days seems to be "decomposition"
multiply the first and last coefficients:
6(-35) = -210
Now look for 2 factors which when multiplied will give you -210 and when added will give you +11
how about 21 and -10
so decompose the middle term of 11x into -10x + 21x
6x^2 - 10x + 21x - 35 = 0
2x(3x - 5) + 7(3x - 5) = 0
(3x-5)(2x + 7) = 0
x = 5/3 or x = -7/2
or
by formula:
x = (-11 ± √(121 - 4(6)(-35)) )/12
= (-11 ± √961)/12
= (-11 ± 31)/12
= 5/3 or -7/2
same as above
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