Solve x² + 12x + 4 = 0
The solutions of x² + 12x + 4 = 0 are x = -6 + 4√2 or x = -6 - 4√2, that is x ≈ -0.34314575 and x ≈ -11.656854.
Change a, b or c (whole numbers, a not 0) and the exact solutions follow.
How to work it out
- 1
Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 1, b = 12, c = 4.
- 2
Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.
- 3
D is positive, so there are two different real solutions.
- 4
The quadratic formula works for every quadratic:
- 5
Simplify (pull the square out of the root) to get the solutions:
- 6
As decimals: x ≈ -0.34314575 and x ≈ -11.656854.
- 7
Check numerically: for x ≈ -0.34314575 the left side is 4.4408921e-15 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ x^2 + 12x + 4 = 0\ \Rightarrow\ x = -6 + 4\sqrt{2},\ -6 - 4\sqrt{2} \]
\[ x^2 + 12x + 4 = 0 \]
\[ D = 12^2 - 4\cdot1\cdot4 = 144 - 16 = 128 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{128}}{2} \]
\[ x = -6 + 4\sqrt{2}\quad\text{or}\quad x = -6 - 4\sqrt{2} \]The parabola y = x² + 12x + 4
| x | y |
|---|---|
| -9 | -23 |
| -8 | -28 |
| -7 | -31 |
| -6 | -32 |
| -5 | -31 |
| -4 | -28 |
| -3 | -23 |
Questions
What are the solutions of x² + 12x + 4 = 0?
The solutions of x² + 12x + 4 = 0 are x = -6 + 4√2 or x = -6 - 4√2, that is x ≈ -0.34314575 and x ≈ -11.656854.
Does x² + 12x + 4 = 0 factor?
Not over the integers, because the discriminant 128 is not a perfect square. Use the quadratic formula.
What is the vertex of y = x² + 12x + 4?
(-6, -32). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -12, product = c/a = 4.