Solve -x² + 12x - 11 = 0
The solutions of -x² + 12x - 11 = 0 are x = 1 or x = 11.
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 = -11.
- 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 = 1 back in: a·x² + b·x + c = -1·(1) + 12·(1) + (-11) = 0 ✓
LaTeX source of the formulas on this page
\[ -x^2 + 12x - 11 = 0\ \Rightarrow\ x = 1,\ 11 \]
\[ -x^2 + 12x - 11 = 0 \]
\[ D = 12^2 - 4\cdot(-1)\cdot(-11) = 144 - 44 = 100 \]
\[ -\left(x - 1\right)\left(x - 11\right) = 0 \]
\[ x - 1 = 0 \Rightarrow x = 1\qquad x - 11 = 0 \Rightarrow x = 11 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{100}}{-2} \]
\[ x = 1\quad\text{or}\quad x = 11 \]The parabola y = -x² + 12x - 11
| x | y |
|---|---|
| 3 | 16 |
| 4 | 21 |
| 5 | 24 |
| 6 | 25 |
| 7 | 24 |
| 8 | 21 |
| 9 | 16 |
Questions
What are the solutions of -x² + 12x - 11 = 0?
The solutions of -x² + 12x - 11 = 0 are x = 1 or x = 11.
Does -x² + 12x - 11 = 0 factor?
Yes: -1(x-1)(x-11) = 0.
What is the vertex of y = -x² + 12x - 11?
(6, 25). 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 = 11.