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Quadratic Equation help?
Hello, ok I have done two simpler equations by myself and I cant to the other two. I have the answers, but need to know step by step how they are gotten.
I must answer it using the -b +- (square root b^2 - 4ac)/2 formula
z^2 - 6z + 7 = 0
and
x^2/3 + 3x - 2 = 0
Thanks again
Ok thanks alot for the help... And the answer to how sharp I am ... well im blunt when it comes to maths after not doing it for 3 years xD I need to learn some for Geology at Uni this year
3 Answers
- RajendiranLv 61 decade agoFavourite answer
1).
z^2 - 6z + 7 = 0
(z - 3)^2 - 2 = 0
z = 3 - √2; z = 1.5858
z = 3 + √2; z = 4.4142
2).
First case:
x^2/3 + 3x - 2 = 0
Multiplying both sides by 3
x^2 + 9x - 6 = 0
(x +9/2)^2 - 105/4 = 0
x = 1/2(-9 - √105); x = -9.6235
x = 1/2(√105 - 9); x = 0.6235
Second case:
x^(2/3) + 3x - 2 = 0
x = 0.466241
- Anonymous1 decade ago
In the first equation a = 1 b = -6 c = 7
b^2 - 4ac = 36 - (4x1x7) = 8
(6 + or - sqrt(8))/2
z = 4.41 or z =1.59
2) a = 1/3 b = 3 c = -2
x = 0.62 or x = -9.62
- Anonymous1 decade ago
Plug in the numbers
-(-6) +/- sqrt[-6^2 - 4(1)(7)]/2(1)
that works down to 6 +/- sqrt(8) / 2
factor out 8 into prime factors of 2 2 2 and you end up with 2(sqrt)2
so now we have 6 +/- 2(sqrt)2 / 2
divide 2 into the integers not under the sqrt and you come up with
3 +/- (sqrt)2
