last one that has a square root equalling zero?

Question

3 + Square root x^ 2 -8x = 0

It looks like squaring the x^2-8x=0
add 3

Answer 1

This is called a Radical Equation.

1. Bring 3 over to the right making it -3.

We have:

sqrt{x^2-8x} = -3

2. Square both sides of the equation.

We now have:

x^2 - 8x = 9

3. Bring 9 over to the left and make it -9.

x^2-8x-9 = 0

4. Factor the trinomial on the left side. A trinomial is a polynomial of THREE terms. See it?

(x - 9) (x+1) = 0

Notice that (x-9) and (x+1) are factors. Two numbers or as in this case, two quantities that are multiplied, are called factors.

5. Set each factor to zero and solve for x.

x - 9 = 0

x = 9

and

x + 1 = 0

x = -1

Our values for x are: 9 and -1.

If you plug these numbers into the orginal radical equation, you will get 3 + square root 9 which does NOT equal zero.

You CANNOT take the square root of a negative number in math.

I say this radical equation has NO SOLUTION.

Guido

Answer 2

I assume you mean 3 + sqrt(x^2 - 8x) = 0 (never hurts to use more parentheses)

Then you have sqrt(x^2 - 8x) = -3, which can only have a solution if the negative square root is intended.

Square both sides: x^2 - 8x = 9, or x^2 - 8x - 9 = 0.

This factors into (x+1)(x-9) = 0, so x=-1 or x=9.

If you plug in either of these values for x, you get 3 + sqrt(9) = 0, which holds if you take the negative square root of 9, which is -3.

Answer 3

This is a trick question The square root of x^2 - 8x would have to come out -3 in order for 3 + that to equal zero. By the definition of principal square root that is impossible.

Answer 4

I suggest you to solve your mathematic problem by ''Newton approximation formula''.