anything squared cannot be negative in the real number set, so your first one has no real number solution
2. correct
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.
4 answers
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 :)
x = [ 0 +/- sqrt(-28) ]/2
x = +/- 2 sqrt -7/2
x = +/- sqrt(-7) or +/- i sqrt 7
=================================
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
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
where i is the sqrt of NEGATVE one