How do you solve #2x^2 + 2x – 1 = 0#?

Question

Eva Adamson · Accepted Answer

#x = (-1 +- sqrt3)/2#

To solve this, we will use the quadratic formula. We know that this is in standard quadratic form, or #ax^2 + bx + c#, where #a = 2#, #b = 2#, and #c = -1#.

The quadratic formula is: #x = (-b +- sqrt(b^2 - 4ac))/(2a)#

Now plug in the values of #a#, #b#, and #c# into the formula to find the value of #x#: #x = (-2 +- sqrt(2^2 - 4(2)(-1)))/(2(2))#

#x = (-2 +- sqrt(4 + 8))/4#

#x = (-2 +- sqrt12)/4#

#x = (-2 +- 2sqrt3)/4#

#x = (-1 +- sqrt3)/2#

I hope this is useful.

Genesis Allen · Answer

undefined To solve the quadratic equation $2x^2 + 2x - 1 = 0$, you can use the quadratic formula, which states that for an equation in the form $ax^2 + bx + c = 0$, the solutions are given by:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, $a = 2$, $b = 2$, and $c = -1$. Plugging these values into the formula gives:

$$x = \frac{-2 \pm \sqrt{2^2 - 4(2)(-1)}}{2(2)}$$
$$x = \frac{-2 \pm \sqrt{4 + 8}}{4}$$
$$x = \frac{-2 \pm \sqrt{12}}{4}$$
$$x = \frac{-2 \pm 2\sqrt{3}}{4}$$
$$x = \frac{-1 \pm \sqrt{3}}{2}$$

So, the solutions to the equation $2x^2 + 2x - 1 = 0$ are $x = \frac{-1 + \sqrt{3}}{2}$ and $x = \frac{-1 - \sqrt{3}}{2}$.