Solve 2x² + 2x − 2 = 0 with the quadratic formula
Quadratic equation, worked out line by line the way a teacher would write it.
Answer
| Solutions | x = (−√5 − 1)/2, x = (√5 − 1)/2 |
Step-by-step solution
9 steps-
1 Given2 x^{2} + 2 x - 2 = 0
Solve for x.
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2 Divide both sides by the common factor 2
Every coefficient is divisible by 2. Dividing both sides by 2 gives smaller numbers and the same solutions, because 0 divided by 2 is still 0.
x^{2} + x - 1 = 0 -
3 This is a quadratic in standard form ax² + bx + c = 0
Every quadratic equation can be arranged as ax² + bx + c = 0. Reading off a = 1, b = 1 and c = -1, signs included, shows which method fits: factoring, taking a square root or the quadratic formula.
a = 1,\quad b = 1,\quad c = -1 -
4 Apply the quadratic formula
The quadratic formula solves any equation ax² + bx + c = 0. It comes from completing the square on the general equation, so it always works, even when the trinomial does not factor.
x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} -
5 Substitute a, b and cx = \frac{-1 \pm \sqrt{1^{2} - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}
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6 Compute the discriminant D = b² − 4ac
D = b² − 4ac is the part under the square root in the formula. If it is positive there are two real solutions, if it is 0 there is one, and if it is negative its square root is not a real number, so there is no real solution.
D = 5Positive discriminant: two distinct real solutions.
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7 Evaluate the square root and the denominatorx = \frac{-1 \pm \sqrt{5}}{2}
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8 Simplifyx = \frac{- \sqrt{5} - 1}{2}\quad \text{or} \quad x = \frac{\sqrt{5} - 1}{2}
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9 Check\begin{aligned}x = \frac{- \sqrt{5} - 1}{2}:\quad 0 = 0\quad\checkmark\\ x = \frac{\sqrt{5} - 1}{2}:\quad 0 = 0\quad\checkmark\end{aligned}
Substituting each solution back makes both sides equal.
Check by substitution
| x | Left side | Right side | |
|---|---|---|---|
| (−√5 − 1)/2 | 0 | 0 | ✓ |
| (√5 − 1)/2 | 0 | 0 | ✓ |
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