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Solve x² + x - 12 = 0

The solutions of x² + x - 12 = 0 are x = 3 or x = -4.

x² +x += 0
x2+x−12=0⇒x=3, −4x^2 + x - 12 = 0 \Rightarrow x = 3,\ -4
Discriminant D = 49 · x ≈ 3, -4

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

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Graph of y = x² + x - 12-6-4-224-15-10-5510152025x = 3x = -4vertexy = x² + x - 12
Graph of y = x² + x - 12. The curve crosses the x-axis at x = 3 and x = -4; the vertex is at (-0.5, -12.25).

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 = 1, c = -12.

    x2+x−12=0x^2 + x - 12 = 0
  2. 2

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

    D=12−4⋅1⋅(−12)=1+48=49D = 1^2 - 4\cdot1\cdot(-12) = 1 + 48 = 49
  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−3)(x+4)=0\left(x - 3\right)\left(x + 4\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−3=0⇒x=3x+4=0⇒x=−4x - 3 = 0 \Rightarrow x = 3\qquad x + 4 = 0 \Rightarrow x = -4
  6. 6

    Method 2, The quadratic formula works for every quadratic:

    x=−b±D2a=−1±492x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{49}}{2}
  7. 7

    Simplify to get the solutions:

    x=3orx=−4x = 3\quad\text{or}\quad x = -4
  8. 8

    Check by putting x = 3 back in: a·x² + b·x + c = 1·(9) + 1·(3) + (-12) = 0 ✓

LaTeX source of the formulas on this page
\[ x^2 + x - 12 = 0\ \Rightarrow\ x = 3,\ -4 \]
\[ x^2 + x - 12 = 0 \]
\[ D = 1^2 - 4\cdot1\cdot(-12) = 1 + 48 = 49 \]
\[ \left(x - 3\right)\left(x + 4\right) = 0 \]
\[ x - 3 = 0 \Rightarrow x = 3\qquad x + 4 = 0 \Rightarrow x = -4 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{49}}{2} \]
\[ x = 3\quad\text{or}\quad x = -4 \]

The parabola y = x² + x - 12

xy
-3.5-3.25
-2.5-8.25
-1.5-11.25
-0.5-12.25
0.5-11.25
1.5-8.25
2.5-3.25

Questions

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

The solutions of x² + x - 12 = 0 are x = 3 or x = -4.

Does x² + x - 12 = 0 factor?

Yes: (x-3)(x+4) = 0.

What is the vertex of y = x² + x - 12?

(-1/2, -49/4). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -1, product = c/a = -12.

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