Solve -x² - 12x = 0
The solutions of -x² - 12x = 0 are x = -12 or x = 0.
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 = 0.
- 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
Method 1, factoring. The roots are rational, so the left side factors:
- 5
A product is zero when one of its factors is zero, so set each factor equal to 0 and solve:
- 6
Method 2, The quadratic formula works for every quadratic:
- 7
Simplify to get the solutions:
- 8
Check by putting x = -12 back in: a·x² + b·x + c = -1·(144) + (-12)·(-12) + 0 = 0 ✓
LaTeX source of the formulas on this page
\[ -x^2 - 12x = 0\ \Rightarrow\ x = -12,\ 0 \]
\[ -x^2 - 12x = 0 \]
\[ D = (-12)^2 - 4\cdot(-1)\cdot0 = 144 + 0 = 144 \]
\[ -\left(x + 12\right)\left(x\right) = 0 \]
\[ x + 12 = 0 \Rightarrow x = -12\qquad x = 0 \Rightarrow x = 0 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{12 \pm \sqrt{144}}{-2} \]
\[ x = -12\quad\text{or}\quad x = 0 \]The parabola y = -x² - 12x
| x | y |
|---|---|
| -9 | 27 |
| -8 | 32 |
| -7 | 35 |
| -6 | 36 |
| -5 | 35 |
| -4 | 32 |
| -3 | 27 |
Questions
What are the solutions of -x² - 12x = 0?
The solutions of -x² - 12x = 0 are x = -12 or x = 0.
Does -x² - 12x = 0 factor?
Yes: -1(x+12)(x) = 0.
What is the vertex of y = -x² - 12x?
(-6, 36). The parabola opens downwards, so this is the maximum.
What are the sum and product of the roots?
Sum = −b/a = -12, product = c/a = 0.