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Solve -x² + 5x - 7 = 0

-x² + 5x - 7 = 0 has no real solutions. The two complex solutions are x = (5 + √3 i)/2 and x = (5 - √3 i)/2.

x² +x += 0
−x2+5x−7=0⇒x=5+3 i2, 5−3 i2 (complex)-x^2 + 5x - 7 = 0 \Rightarrow x = \frac{5 + \sqrt{3}\, i}{2},\ \frac{5 - \sqrt{3}\, i}{2}\ \text{(complex)}
Discriminant D = -3

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

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Graph of y = -x² + 5x - 7-1123456-16-14-12-10-8-6-4-2vertexy = -x² + 5x - 7
Graph of y = -x² + 5x - 7. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (2.5, -0.75).

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 = 5, c = -7.

    −x2+5x−7=0-x^2 + 5x - 7 = 0
  2. 2

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

    D=52−4⋅(−1)⋅(−7)=25−28=−3D = 5^2 - 4\cdot(-1)\cdot(-7) = 25 - 28 = -3
  3. 3

    D is negative, so there are no real solutions; the two solutions are complex numbers.

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=−5±−3−2x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{-3}}{-2}
  5. 5

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

    x=5+3 i2orx=5−3 i2x = \frac{5 + \sqrt{3}\, i}{2}\quad\text{or}\quad x = \frac{5 - \sqrt{3}\, i}{2}
LaTeX source of the formulas on this page
\[ -x^2 + 5x - 7 = 0\ \Rightarrow\ x = \frac{5 + \sqrt{3}\, i}{2},\ \frac{5 - \sqrt{3}\, i}{2} \]
\[ -x^2 + 5x - 7 = 0 \]
\[ D = 5^2 - 4\cdot(-1)\cdot(-7) = 25 - 28 = -3 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{-3}}{-2} \]
\[ x = \frac{5 + \sqrt{3}\, i}{2}\quad\text{or}\quad x = \frac{5 - \sqrt{3}\, i}{2} \]

The parabola y = -x² + 5x - 7

xy
-0.5-9.75
0.5-4.75
1.5-1.75
2.5-0.75
3.5-1.75
4.5-4.75
5.5-9.75

Questions

What are the solutions of -x² + 5x - 7 = 0?

-x² + 5x - 7 = 0 has no real solutions. The two complex solutions are x = (5 + √3 i)/2 and x = (5 - √3 i)/2.

Does -x² + 5x - 7 = 0 factor?

Not over the integers, because the discriminant -3 is negative. Use the quadratic formula.

What is the vertex of y = -x² + 5x - 7?

(5/2, -3/4). The parabola opens downwards, so this is the maximum.

What are the sum and product of the roots?

Sum = −b/a = 5, product = c/a = 7.

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