Since the discriminant is positive, we can conclude that the equation has two real solutions. :ĭetermine the type and number of solutions: x 2 − 2 x − 4 = 0. Given a x 2 + b x + c = 0, where a, b, and c are real numbers and a ≠ 0, the solutions can be calculated using the quadratic formula The formula x = − b ± b 2 − 4 a c 2 a, which gives the solutions to any quadratic equation in the standard form a x 2 + b x + c = 0, where a, b, and c are real numbers and a ≠ 0. This derivation gives us a formula that solves any quadratic equation in standard form. ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 2 a x = − b 2 a ± b 2 − 4 a c 2 a x = − b ± b 2 − 4 a c 2 a X 2 + b a x + b 2 4 a 2 = − c a + b 2 4 a 2 ( x + b 2 a ) ( x + b 2 a ) = − c a + b 2 4 a 2 ( x + b 2 a ) 2 = − 4 a c 4 a 2 + b 2 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x 2 + b a x = − c aĭetermine the constant that completes the square: take the coefficient of x, divide it by 2, and then square it.Īdd this to both sides of the equation to complete the square and then factor. x 2 + b a x + c a = 0 S u b t r a c t c a f r o m b o t h s i d e s. a x 2 + b x + c a = 0 a D i v i d e b o t h s i d e s b y a. Here a, b, and c are real numbers and a ≠ 0:Ī x 2 + b x + c = 0 S t a n d a r d f o r m o f a q u a d r a t i c e q u a t i o n. To do this, we begin with a general quadratic equation in standard form and solve for x by completing the square. In this section, we will develop a formula that gives the solutions to any quadratic equation in standard form.
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