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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, −12x^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-14-12-10-8-6-4-22-40-30-20-1010203040x = 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=0x^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\cdot1\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±1442x = \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
\[ x^2 + 12x = 0\ \Rightarrow\ x = 0,\ -12 \]
\[ x^2 + 12x = 0 \]
\[ D = 12^2 - 4\cdot1\cdot0 = 144 + 0 = 144 \]
\[ \left(x\right)\left(x + 12\right) = 0 \]
\[ x = 0 \Rightarrow x = 0\qquad x + 12 = 0 \Rightarrow x = -12 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{144}}{2} \]
\[ x = 0\quad\text{or}\quad x = -12 \]

The parabola y = x² + 12x

xy
-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 = 0 or x = -12.

Does x² + 12x = 0 factor?

Yes: (x)(x+12) = 0.

What is the vertex of y = x² + 12x?

(-6, -36). 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 = 0.

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