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Solve -x² + 12x = 0

The solutions of -x² + 12x = 0 are x = 0 or x = 12.

x² +x += 0
−x2+12x=0⇒x=0, 12-x^2 + 12x = 0 \Rightarrow x = 0,\ 12
Discriminant D = 144 · x ≈ 0, 12

Change a, b or c (whole numbers, a not 0) and the exact solutions follow.

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Graph of y = -x² + 12x-22468101214-40-30-20-1010203040x = 0x = 12vertexy = -x² + 12x
Graph of y = -x² + 12x. The curve crosses the x-axis at x = 0 and x = 12; the vertex is at (6, 36).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = -1, b = 12, c = 0.

    −x2+12x=0-x^2 + 12x = 0
  2. 2

    Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.

    D=122−4⋅(−1)⋅0=144+0=144D = 12^2 - 4\cdot(-1)\cdot0 = 144 + 0 = 144
  3. 3

    D is positive, so there are two different real solutions.

  4. 4

    Method 1, factoring. The roots are rational, so the left side factors:

    −(x)(x−12)=0-\left(x\right)\left(x - 12\right) = 0
  5. 5

    A product is zero when one of its factors is zero, so set each factor equal to 0 and solve:

    x=0⇒x=0x−12=0⇒x=12x = 0 \Rightarrow x = 0\qquad x - 12 = 0 \Rightarrow x = 12
  6. 6

    Method 2, The quadratic formula works for every quadratic:

    x=−b±D2a=−12±144−2x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{144}}{-2}
  7. 7

    Simplify to get the solutions:

    x=0orx=12x = 0\quad\text{or}\quad x = 12
  8. 8

    Check by putting x = 0 back in: a·x² + b·x + c = -1·(0) + 12·(0) + 0 = 0 ✓

LaTeX source of the formulas on this page
\[ -x^2 + 12x = 0\ \Rightarrow\ x = 0,\ 12 \]
\[ -x^2 + 12x = 0 \]
\[ D = 12^2 - 4\cdot(-1)\cdot0 = 144 + 0 = 144 \]
\[ -\left(x\right)\left(x - 12\right) = 0 \]
\[ x = 0 \Rightarrow x = 0\qquad x - 12 = 0 \Rightarrow x = 12 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{144}}{-2} \]
\[ x = 0\quad\text{or}\quad x = 12 \]

The parabola y = -x² + 12x

xy
327
432
535
636
735
832
927

Questions

What are the solutions of -x² + 12x = 0?

The solutions of -x² + 12x = 0 are x = 0 or x = 12.

Does -x² + 12x = 0 factor?

Yes: -1(x)(x-12) = 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.

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