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.
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
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
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
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
To solve the equation -x^4 + 200 = 102x^2, you correctly rearranged it to x^4 + 102x^2 - 200 = 0. However, to further simplify the equation, you need to bring all terms to one side to make it equal to zero.
So, the equation becomes x^4 + 102x^2 - 200 = 0.
To solve this equation, you can use the quadratic formula. However, since the equation is in terms of x^2, you can introduce a substitution to simplify it even further.
Let's substitute x^2 = y:
Then, the equation becomes y^2 + 102y - 200 = 0.
Now, you can use the quadratic formula to find the values of y:
y = (-b ± sqrt(b^2 - 4ac)) / (2a),
where a = 1, b = 102, and c = -200.
Plugging in these values, we get:
y = (-102 ± sqrt(102^2 - 4*1*(-200))) / (2*1),
Simplifying further:
y = (-102 ± sqrt(10404 + 800)) / 2,
y = (-102 ± sqrt(11204)) / 2,
y = (-102 ± 105.92) / 2.
This gives us two possible values of y:
y = (-102 + 105.92) / 2 = 1.96,
y = (-102 - 105.92) / 2 = -103.96.
Now, we substitute y back into x^2 to get the values of x:
For y = 1.96, x^2 = 1.96,
Taking the square root of both sides, x ≈ ±1.4.
For y = -103.96, x^2 = -103.96,
This equation has no real solutions because you cannot take the square root of a negative number.
Therefore, the solutions to the equation are approximately x = ±1.4.
Regarding the second problem, to find the maximum height of the arch represented by h(x) = -0.03x^2 + 3x, you need to find the vertex of the parabolic function. The vertex represents the point at which the height is maximized.
To find the vertex, you can use the formula x = -b / 2a, where a = -0.03 and b = 3. Applying the formula:
x = -3 / (2 * -0.03) = -3 / -0.06 = 50,
This means that the maximum height of the arch occurs when x = 50 meters.
Now, you can substitute x = 50 back into the equation h(x) = -0.03x^2 + 3x to find the maximum height:
h(50) = -0.03 * (50^2) + 3 * 50 = -0.03 * 2500 + 150 = -75 + 150 = 75 meters.
Thus, the maximum height of the arch is 75 meters.