Algebra 2 question help?!
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Algebra 2 question help?!

[From: ] [author: ] [Date: 12-10-07] [Hit: ]
Expand it and you get x^2 - 10x + 37. Look at the discriminant, it is -48. This quadratic has two complex solutions which you can find by using the quadratic formula. But there are no zeros in this equation which makes sense since this quadratic function has been shifted right 5 units and then moved up 12 units, you expect it to have no zeros.......
Find the zeros of each function by using the Quadratic Formula.
Write it out step by step, please!

h(x) =(x-5)2+12

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= x^2-10x+25+12
x^2-10x+37

= 10 +- square root of {(10)^2-4(1)(37)}/2
= 10 +- [100-148]/2
= 10 +- SR [-48]/2
= 10 +- 4 SR [3]/2
= 5 +- 2 SR [3]

Sr is square root so both answers would be 5 plus or minus 2 square root of 3
Hope that helped :)

-
h(x)=(x-5)^2+12
Expand it and you get x^2 - 10x + 37. Look at the discriminant, it is -48. This quadratic has two complex solutions which you can find by using the quadratic formula. But there are no zeros in this equation which makes sense since this quadratic function has been shifted right 5 units and then moved up 12 units, you expect it to have no zeros. Using the quadratic formula, you get:
[10+/-sqrt(-48)]/2.

Simplifying that, the two complex solutions are:
x = 5-2 i sqrt(3)
x = 5+2 i sqrt(3)

Hope I helped.........................!!
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