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Solve x² - 12x + 4 = 0

The solutions of x² - 12x + 4 = 0 are x = 6 + 4√2 or x = 6 - 4√2, that is x ≈ 11.656854 and x ≈ 0.34314575.

x² +x += 0
x2−12x+4=0⇒x=6+42, 6−42x^2 - 12x + 4 = 0 \Rightarrow x = 6 + 4\sqrt{2},\ 6 - 4\sqrt{2}
Discriminant D = 128 · x ≈ 11.6569, 0.343146

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

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Graph of y = x² - 12x + 4-22468101214-40-30-20-1010203040x = 11.656854x = 0.34314575vertexy = x² - 12x + 4
Graph of y = x² - 12x + 4. The curve crosses the x-axis at x = 11.656854 and x = 0.34314575; the vertex is at (6, -32).

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 = 4.

    x2−12x+4=0x^2 - 12x + 4 = 0
  2. 2

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

    D=(−12)2−4⋅1⋅4=144−16=128D = (-12)^2 - 4\cdot1\cdot4 = 144 - 16 = 128
  3. 3

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

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=12±1282x = \frac{-b \pm \sqrt{D}}{2a} = \frac{12 \pm \sqrt{128}}{2}
  5. 5

    Simplify (pull the square out of the root) to get the solutions:

    x=6+42orx=6−42x = 6 + 4\sqrt{2}\quad\text{or}\quad x = 6 - 4\sqrt{2}
  6. 6

    As decimals: x ≈ 11.656854 and x ≈ 0.34314575.

  7. 7

    Check numerically: for x ≈ 11.656854 the left side is 0 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ x^2 - 12x + 4 = 0\ \Rightarrow\ x = 6 + 4\sqrt{2},\ 6 - 4\sqrt{2} \]
\[ x^2 - 12x + 4 = 0 \]
\[ D = (-12)^2 - 4\cdot1\cdot4 = 144 - 16 = 128 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{12 \pm \sqrt{128}}{2} \]
\[ x = 6 + 4\sqrt{2}\quad\text{or}\quad x = 6 - 4\sqrt{2} \]

The parabola y = x² - 12x + 4

xy
3-23
4-28
5-31
6-32
7-31
8-28
9-23

Questions

What are the solutions of x² - 12x + 4 = 0?

The solutions of x² - 12x + 4 = 0 are x = 6 + 4√2 or x = 6 - 4√2, that is x ≈ 11.656854 and x ≈ 0.34314575.

Does x² - 12x + 4 = 0 factor?

Not over the integers, because the discriminant 128 is not a perfect square. Use the quadratic formula.

What is the vertex of y = x² - 12x + 4?

(6, -32). 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 = 4.

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