How do you find the zeros of $f (x) = 2 x^{2} + x - 3$ $f(x)=2x^2+x-3$ ?

Question

11310 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-05T15:42:01

$x_{1} = 1 and x_{2} = - \frac{3}{2}$ $x_1=1 and x_2=-3/2$

Factor the equation $f (x)$ $f(x)$ (I'll use decomposition),

$f (x) = 2 x^{2} + 2 x - 3 x - 3$ $f(x)=2x^2+2x-3x-3$

Factor,

$f (x) = 2 x (x - 1) + 3 (x - 1)$ $f(x)=2x(x-1)+3(x-1)$

Simplify,

$f (x) = (2 x + 3) (x - 1)$ $f(x)=(2x+3)(x-1)$

Let $f (x) = 0,$ $f(x)=0,$

$0 = (2 x + 3) (x - 1)$ $0=(2x+3)(x-1)$

The values of x that make this statement above true (zeros), are

$x_{1} = 1 and x_{2} = - \frac{3}{2}$ $x_1=1 and x_2=-3/2$

How do you find the zeros of f(x)=2x2+x−3f(x)=2x^2+x-3?