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Solve -x² + 6x + 4 = 0

The solutions of -x² + 6x + 4 = 0 are x = 3 + √13 or x = 3 - √13, that is x ≈ -0.60555128 and x ≈ 6.6055513.

x² +x += 0
−x2+6x+4=0⇒x=3+13, 3−13-x^2 + 6x + 4 = 0 \Rightarrow x = 3 + \sqrt{13},\ 3 - \sqrt{13}
Discriminant D = 52 · x ≈ -0.605551, 6.60555

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

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Graph of y = -x² + 6x + 4-22468-25-20-15-10-551015x = -0.60555128x = 6.6055513vertexy = -x² + 6x + 4
Graph of y = -x² + 6x + 4. The curve crosses the x-axis at x = -0.60555128 and x = 6.6055513; the vertex is at (3, 13).

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 = 6, c = 4.

    −x2+6x+4=0-x^2 + 6x + 4 = 0
  2. 2

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

    D=62−4⋅(−1)⋅4=36+16=52D = 6^2 - 4\cdot(-1)\cdot4 = 36 + 16 = 52
  3. 3

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

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=−6±52−2x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{52}}{-2}
  5. 5

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

    x=3+13orx=3−13x = 3 + \sqrt{13}\quad\text{or}\quad x = 3 - \sqrt{13}
  6. 6

    As decimals: x ≈ -0.60555128 and x ≈ 6.6055513.

  7. 7

    Check numerically: for x ≈ -0.60555128 the left side is 1.3322676e-15 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ -x^2 + 6x + 4 = 0\ \Rightarrow\ x = 3 + \sqrt{13},\ 3 - \sqrt{13} \]
\[ -x^2 + 6x + 4 = 0 \]
\[ D = 6^2 - 4\cdot(-1)\cdot4 = 36 + 16 = 52 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{52}}{-2} \]
\[ x = 3 + \sqrt{13}\quad\text{or}\quad x = 3 - \sqrt{13} \]

The parabola y = -x² + 6x + 4

xy
04
19
212
313
412
59
64

Questions

What are the solutions of -x² + 6x + 4 = 0?

The solutions of -x² + 6x + 4 = 0 are x = 3 + √13 or x = 3 - √13, that is x ≈ -0.60555128 and x ≈ 6.6055513.

Does -x² + 6x + 4 = 0 factor?

Not over the integers, because the discriminant 52 is not a perfect square. Use the quadratic formula.

What is the vertex of y = -x² + 6x + 4?

(3, 13). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 6, product = c/a = -4.

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    Solve -x² + 6x + 4 = 0: x = 3 + √13 or x = 3 - √13 (steps) | WordGPT