Write each equation in standard form. Solve by using the Quadratic Formula.
1. -7 = x^2
A: ?
2. x^2 - 16 = 0
A: x = 4 or x = -4.
anything squared cannot be negative in the real number set, so your first one has no real number solution
2. correct
x^2 + 0 x + 7 = 0
x = [ 0 +/- sqrt(-28) ]/2
x = +/- 2 sqrt -7/2
x = +/- sqrt(-7) or +/- i sqrt 7
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x^2 - 0 x -16 = 0
x = [ 0 + /- sqrt (0 +4*16) ]/2
x = +/- 2*4/2
x = +/- 4
what else is new :)
Are you sure? I haven't written the equation in standard form yet.
1. x^2 + 0x + 7 = 0
Yes, the only solutions to the first problem are the imaginary numbers +/- i sqrt 7
where i is the sqrt of NEGATVE one
To write each equation in standard form, we need to rearrange the terms so that the equation is in the form ax^2 + bx + c = 0.
1. Starting with the equation -7 = x^2, we can simply rearrange the terms, so it becomes x^2 + 7 = 0.
Now, we can solve it using the quadratic formula.
The quadratic formula states that for an equation ax^2 + bx + c = 0, the solutions x can be found using the formula:
x = (-b ± sqrt(b^2 - 4ac))/(2a)
In this case, a = 1, b = 0, and c = 7. Plugging in these values into the formula, we get:
x = (-0 ± sqrt(0^2 - 4(1)(7)))/(2(1))
= (± sqrt(-28))/2
Since we have a negative value under the square root, it means the equation has no real solutions.
Therefore, there are no real values of x that satisfy the equation.
2. For the equation x^2 - 16 = 0, we already have it in the form ax^2 + bx + c = 0.
Now, applying the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac))/(2a)
Here, a = 1, b = 0, and c = -16. Plugging in these values into the formula, we get:
x = (± sqrt(0^2 - 4(1)(-16)))/(2(1))
= (± sqrt(64))/2
= (± 8)/2
Simplifying further, we get:
x = 8/2 or x = -8/2
x = 4 or x = -4
Therefore, the solutions to the equation x^2 - 16 = 0 are x = 4 and x = -4.