How do you solve $2 x^{2} - 3 x + 1 = 0$ $2x^2 - 3x + 1 = 0$ ?

Question

How do you solve $2 x^{2} - 3 x + 1 = 0$ $2x^2 - 3x + 1 = 0$ ?

1 Answer

Impact of this question

6420 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-04T04:45:35

$x = \frac{1}{2}, 1$ $x = 1/2, 1$

Here's how I solved it:

Explanation:

You solve this by using the quadratic formula. This equation is written in standard form, or $a x^{2} + b x + c = 0$ $ax^2 + bx + c = 0$ . The quadratic formula is $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2-4ac))/(2a)$ .

So let's plug those numbers in and solve for $x$ $x$ :
$x = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 (2) (1)}}{2 (2)}$ $x = (-(-3) +- sqrt((-3)^2 - 4(2)(1)))/(2(2))$

$x = \frac{3 \pm \sqrt{9 - 8}}{4}$ $x = (3 +- sqrt(9 - 8))/4$

$x = \frac{3 \pm \sqrt{1}}{4}$ $x = (3 +- sqrt(1))/4$

$x = \frac{3 \pm 1}{4}$ $x = (3 +- 1)/4$

So $x = \frac{3 + 1}{4}$ $x = (3+1)/4$ and $\frac{3 - 1}{4}$ $(3-1)/4$ .

$x = \frac{4}{4} \to x = 1$ $x = 4/4 -> x = 1$

$x = \frac{2}{4} \to x = \frac{1}{2}$ $x = 2/4 -> x = 1/2$

$x = \frac{1}{2}, 1$ $x = 1/2, 1$

Hope this helps!

Science

Math

Humanities

... and beyond

How do you solve $2 x^{2} - 3 x + 1 = 0$ $2x^2 - 3x + 1 = 0$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 2x^2 - 3x + 1 = 02x2−3x+1=02x^2 - 3x + 1 = 0?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $2 x^{2} - 3 x + 1 = 0$ $2x^2 - 3x + 1 = 0$ ?