blank spaces are a (+) sign and (=)
ax2 + bx + c = 0 , find the value of x . b2-4ac>o x is real (2 different values will solve) b2-4ac=o -> a double root (a single real number will solve it) x=real numbers. b2-4ac<0 x= two complex number roots (either pure imaginary or a complex number with real and imaginary components)
If ax2+bx+c=0, then x=(-b/2a)± [√(b2-4ac)]/2a
A quadratic involving x and y is usually in the form 'y = ax2 + bx + c'. This form is y in terms of x, so we must rearrange it. y = ax2 + bx + c y/a = x2 + bx/a + c/a y/a = x2 + bx/a + d + e, where c/a = d + e, e = (b/a)2 y/a - e = x2 + bx/a + d y/a -e = (x + b/a)2 √(y/a - e) = x + b/a √(y/a - e) - b/a = x
Two: one is 0, the other is -b/a ax2 + bx + c = 0, but c = 0 ⇒ ax2 + bx + 0 = 0 ⇒ ax2 + bx = 0 ⇒ x(ax + b) = 0 ⇒ x = 0 or (ax + b) = 0 ⇒ x = -b/a
How you solve an equation that doesn't factor is to plug a quadratic equation's format; ax2+bx+c into the quadratic formula which is x=-b+square root to (b2-4ac)/2a.
y = ax2 + bx + c then you have this equation to solve, where y =0. The variables are referenced from the equation above. x = {-b ± √( b2- 4ac)}/ 2a
put it into the quadratic formula: for ax2+bx+c=0 x= - b (plus or minus)rad(b2-4ac) 2a
where ax2 + bx + c = 0 x = [-b +/- √(b2 -4ac)] / 2a
A quadratic function is ax2+bx+c You can solve for x by using the quadratic formula, which, as the formula requires the use of square roots, would be tricky to put here.
If you are using square roots, the simplest way of solving: ax2 + bx + c = 0 is x = [-b ± sqrt(b2-4ac)]/(2a)
a = bx + c bx + (c-a) = 0 x = (a-c)/b It's simple because there is only one solution. This is a quadratic equation: ax2 + bx + c = 0 It has two solutions.
The coefficient of x