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Solve -x² + x + 4 = 0

The solutions of -x² + x + 4 = 0 are x = (1 + √17)/2 or x = (1 - √17)/2, that is x ≈ -1.5615528 and x ≈ 2.5615528.

x² +x += 0
−x2+x+4=0⇒x=1+172, 1−172-x^2 + x + 4 = 0 \Rightarrow x = \frac{1 + \sqrt{17}}{2},\ \frac{1 - \sqrt{17}}{2}
Discriminant D = 17 · x ≈ -1.56155, 2.56155

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

Graph of y = -x² + x + 4-4-3-2-112345-15-10-55x = -1.5615528x = 2.5615528vertexy = -x² + x + 4
Graph of y = -x² + x + 4. The curve crosses the x-axis at x = -1.5615528 and x = 2.5615528; the vertex is at (0.5, 4.25).

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 = 4.

    −x2+x+4=0-x^2 + x + 4 = 0
  2. 2

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

    D=12−4⋅(−1)⋅4=1+16=17D = 1^2 - 4\cdot(-1)\cdot4 = 1 + 16 = 17
  3. 3

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

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=−1±17−2x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{17}}{-2}
  5. 5

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

    x=1+172orx=1−172x = \frac{1 + \sqrt{17}}{2}\quad\text{or}\quad x = \frac{1 - \sqrt{17}}{2}
  6. 6

    As decimals: x ≈ -1.5615528 and x ≈ 2.5615528.

  7. 7

    Check numerically: for x ≈ -1.5615528 the left side is 0 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ -x^2 + x + 4 = 0\ \Rightarrow\ x = \frac{1 + \sqrt{17}}{2},\ \frac{1 - \sqrt{17}}{2} \]
\[ -x^2 + x + 4 = 0 \]
\[ D = 1^2 - 4\cdot(-1)\cdot4 = 1 + 16 = 17 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{17}}{-2} \]
\[ x = \frac{1 + \sqrt{17}}{2}\quad\text{or}\quad x = \frac{1 - \sqrt{17}}{2} \]

The parabola y = -x² + x + 4

xy
-2.5-4.75
-1.50.25
-0.53.25
0.54.25
1.53.25
2.50.25
3.5-4.75

Questions

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

The solutions of -x² + x + 4 = 0 are x = (1 + √17)/2 or x = (1 - √17)/2, that is x ≈ -1.5615528 and x ≈ 2.5615528.

Does -x² + x + 4 = 0 factor?

Not over the integers, because the discriminant 17 is not a perfect square. Use the quadratic formula.

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

(1/2, 17/4). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 1, product = c/a = -4.

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