Approximate to the nearest tenth the positive real zeroes of f(x)=x^2+1. I got .6
Solve over the set of complex numbers.
x^4-13x^2+36=0 i got -3,-2,2,3 but i don't know if i even did it right.
Solve x^2+4x9=0 by completing the square. I don't even know where to begin on this one. So help me please!!!!! I have a test on this stuff Wed.
Huh? x^2 + 1 = 0 has no real zeros
x = + or - sqrt (-1) = +i or -i
let y = x^2
then
y^2 - 13 y + 36 = 0
I will do this by completing the square so you see how
y^2 - 13 y = -36
take half of 13, sqare it, add to both sides
y^2 - 13 y +169/4 = -36 + 169/4
(y-6.5)^2 = 6.25
y - 6.25 = +2.5 or y - 6.25 = -2.5
so
y = 8.75 or y = 3.75
BUT y = x ^2
so
x = + sqrt(8.75)
or
x = -sqrt (8.75)
or
x = + sqrt (3.75)
or
x = -sqrt(3.75)
x^2+4x9=0 This has a typo but I already showed you how to complete square above.
One thing is that coef of x^2 must be one. In your problems it already is, but if it were not, you would have to divide both sides by the non-zero coefficient of x^2 before starting.
To find the approximate value of the positive real zeros of the function f(x) = x^2 + 1, you can set the function equal to zero and solve for x:
x^2 + 1 = 0
Since the equation does not have any real solutions, you will not be able to approximate the positive real zeros. The graph of the function (a parabola) does not intersect the x-axis.
For the second question, to solve the equation x^4 - 13x^2 + 36 = 0, you can use factoring or the quadratic formula.
Using factoring, you can rewrite the equation as:
(x^2 - 9)(x^2 - 4) = 0
Now you have two equations to solve:
x^2 - 9 = 0 and x^2 - 4 = 0
For the first equation, you can take the square root of both sides:
x^2 = 9
x = ±3
For the second equation, again, you can take the square root of both sides:
x^2 = 4
x = ±2
So the solutions to x^4 - 13x^2 + 36 = 0 over the set of complex numbers are x = -3, -2, 2, 3.
Lastly, for the equation x^2 + 4x + 9 = 0, you can solve it by completing the square. Here's how:
Move the constant term to the right side:
x^2 + 4x = -9
Take half of the coefficient of x (which is 2), square it (which is 4), and add it to both sides of the equation:
x^2 + 4x + 4 = -9 + 4
(x + 2)^2 = -5
Now take the square root of both sides:
x + 2 = ±√(-5)
Since the square root of a negative number is not a real number, the equation has no real solutions. It has two complex solutions:
x = -2 + i√5 and x = -2 - i√5.
Remember to always double-check your work and go over examples in your textbook or class notes to ensure you understand how to solve these types of problems for your upcoming test.