ax2 + c = 0
Subtract c from both sides: ax2 = -c
divide both sides by a: x2 = -c/a
and so x = ± sqrt(-c/a)
This has real solutions only if -c/a ≥ 0, that is, if c/a ≤ 0. Depending on the level you are at, if c/a ≥ 0 then there are the imaginary solutions i*sqrt(c/a).
104
Since a squared plus b squared equals c squared, that is the same as c equals the square root of a squared plus b squared. This can be taken into squaring and square roots to infinity and still equal c, as long as there is the same number of squaring and square roots in the problem. Since this question asks for a and b squared three times, and also three square roots of a and b both, they equal c. Basically, they cancel each other out.
89
pythagoras
b= 10
X = √63
b = 14324.80366
2 squared plus 2 x 3 = 10, 7 squared plus 7 x 2 = 63, 6 squared plus 6 x 5 = 66,8 squared plus 8 x 4 = 96 so 9 squared plus 9 x 7 = 81 + 63 = 144.
16
This is the common form of the Pythagorean Theorem. It describes the relationship between the two legs of a right triangle and the hypotenuse.
4
a=b=3.60555