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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 ≈ 0.34314575 and x ≈ 11.656854.

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

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

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Graph of y = -x² + 12x - 4-22468101214-30-20-1010203040x = 0.34314575x = 11.656854vertexy = -x² + 12x - 4
Graph of y = -x² + 12x - 4. The curve crosses the x-axis at x = 0.34314575 and x = 11.656854; 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=0-x^2 + 12x - 4 = 0
  2. 2

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

    D=122−4⋅(−1)⋅(−4)=144−16=128D = 12^2 - 4\cdot(-1)\cdot(-4) = 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±128−2x = \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 ≈ 0.34314575 and x ≈ 11.656854.

  7. 7

    Check numerically: for x ≈ 0.34314575 the left side is -4.4408921e-15 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ -x^2 + 12x - 4 = 0\ \Rightarrow\ x = 6 + 4\sqrt{2},\ 6 - 4\sqrt{2} \]
\[ -x^2 + 12x - 4 = 0 \]
\[ D = 12^2 - 4\cdot(-1)\cdot(-4) = 144 - 16 = 128 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm \sqrt{128}}{-2} \]
\[ x = 6 + 4\sqrt{2}\quad\text{or}\quad x = 6 - 4\sqrt{2} \]

The parabola y = -x² + 12x - 4

xy
323
428
531
632
731
828
923

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 ≈ 0.34314575 and x ≈ 11.656854.

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 downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 12, product = c/a = 4.

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    Solve -x² + 12x - 4 = 0: x = 6 + 4√2 or x = 6 - 4√2 (steps) | WordGPT