Posted by **Lucy** on Saturday, February 23, 2008 at 5:11pm.

The question simply states-solve.

-x^4+200=102x^2

I rearranged the equation so that I have x^-102x^2+200=0

Since I cannot factor this my solution is x= -51 =/-2 square root of 2801 divided by 2. Is this correct.

Also a second problem that I do not know how to work is this:

The shape of a supporting arch can be modeled by h(x)=-0.03x^2+3x, where h(x) represents the height of the arch and x represents the height of the arch and x represents the horizontal distance from one end of the base tothe arch in meters. Find the maximum height if the arch.

Any help on how to start this problem would be helpful.

Thanks.

- Algebra II -
**Reiny**, Saturday, February 23, 2008 at 5:25pm
but it does factor to

(x^2 - 100)(x^2 - 2) = 0

so x = ± 10 or x = ± √2

BTW, check your typing

in the first you have a negative in front,

in the second you don't have an exponent on your first term

I assumed it was x^4 - 102x^2 + 200 = 0

- Algebra II -
**drwls**, Saturday, February 23, 2008 at 5:32pm
It looks to me like the rearranged equation should be

x^4 + 102 x^2 -200 = 0, That is not what you wrote.

The answer is x2 = -51 +/- 52.924,

= 1.924 or -103.924

The only real solution for x is +/- sqrt 1.924

= +/- 1.387

In your second problem, the highest point is where

dh/dx = -.06x + 3 = 0

It occurs at x = 60 m

The height there is -.03*60^2 + 3*60 = 72 m

- Algebra II -
**Reiny**, Saturday, February 23, 2008 at 5:35pm
For the second one, I am going to use Calculus

h(x) = -.03x^2 + 3x

h'(x) - -.06x + 3

= 0 for a max of h

-.06x = -3

x = 50

so h(50) = -.03(2500) + 3(50)

=75

- Algebra II -
**drwls**, Saturday, February 23, 2008 at 5:47pm
I made a math error on the second one. I can't even punch a calculator anymore. Go with Reiny's answer. On the first problem, I assumed your first equation was correctly typed. If the minus sign should not have been in front of x^4, as Reiny assumed, you get an easly factored equation and Reiny's answers

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