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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, −3−13x^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.

Graph of y = x² + 6x - 4-8-6-4-22-101020x = 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=0x^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\cdot1\cdot(-4) = 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±522x = \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
\[ x^2 + 6x - 4 = 0\ \Rightarrow\ x = -3 + \sqrt{13},\ -3 - \sqrt{13} \]
\[ x^2 + 6x - 4 = 0 \]
\[ D = 6^2 - 4\cdot1\cdot(-4) = 36 + 16 = 52 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{52}}{2} \]
\[ x = -3 + \sqrt{13}\quad\text{or}\quad x = -3 - \sqrt{13} \]

The parabola y = x² + 6x - 4

xy
-6-4
-5-9
-4-12
-3-13
-2-12
-1-9
0-4

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 upwards, so this is the minimum.

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