That factors to (x + 1)(x - 3)
x = -1, 3
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.
yazen is so cool
Using the quadratic formula-- ((negative b plus or minus the square root of b squared minus 4ac) divided by (2a)) you'll want to google that so you can see it in numerical form. a, b, and c are the coefficiants of your three terms ( 2 is a, -5 is b, and 2 is c) The answer is (x-2)(2x-1).
25
It's impossible. Powers are always either positive or complex.
It is: 9-4 = 5
(-x)2 = 3=> -x = ±√3=> x = -√3 and x = √3
add 3 to both sides x2 -3(+3) =162 (+3) x2 = 165 x ~ +- 12.84
3x^2 + 5x - 8 = 0 + 8 + 8 3x^2 + 5x = 0 8x ^2 = 0
By factoring I get x-3 divided by x+3
3x2 - 11x - 4 = 0(3x + 1)(x - 4)x = -1/3 and 4
If you are looking for the zeros of this function: x = -2 plus or minus 2 X the square root of 5.
5
Forty-three minus nine equals 34.
Add 2.9 to both sides. y = 8.2
Three fourths minus three sixths equals one fourth
If you mean: 10ac -x/11 = -3 then a as the subject is a = (-33+x)/110c