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Solve -x² + 5x = 0

The solutions of -x² + 5x = 0 are x = 0 or x = 5.

x² +x += 0
−x2+5x=0⇒x=0, 5-x^2 + 5x = 0 \Rightarrow x = 0,\ 5
Discriminant D = 25 · x ≈ 0, 5

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

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Graph of y = -x² + 5x-2-11234567-20-15-10-55x = 0x = 5vertexy = -x² + 5x
Graph of y = -x² + 5x. The curve crosses the x-axis at x = 0 and x = 5; the vertex is at (2.5, 6.25).

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 = 5, c = 0.

    −x2+5x=0-x^2 + 5x = 0
  2. 2

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

    D=52−4⋅(−1)⋅0=25+0=25D = 5^2 - 4\cdot(-1)\cdot0 = 25 + 0 = 25
  3. 3

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

  4. 4

    Method 1, factoring. The roots are rational, so the left side factors:

    −(x)(x−5)=0-\left(x\right)\left(x - 5\right) = 0
  5. 5

    A product is zero when one of its factors is zero, so set each factor equal to 0 and solve:

    x=0⇒x=0x−5=0⇒x=5x = 0 \Rightarrow x = 0\qquad x - 5 = 0 \Rightarrow x = 5
  6. 6

    Method 2, The quadratic formula works for every quadratic:

    x=−b±D2a=−5±25−2x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{25}}{-2}
  7. 7

    Simplify to get the solutions:

    x=0orx=5x = 0\quad\text{or}\quad x = 5
  8. 8

    Check by putting x = 0 back in: a·x² + b·x + c = -1·(0) + 5·(0) + 0 = 0 ✓

LaTeX source of the formulas on this page
\[ -x^2 + 5x = 0\ \Rightarrow\ x = 0,\ 5 \]
\[ -x^2 + 5x = 0 \]
\[ D = 5^2 - 4\cdot(-1)\cdot0 = 25 + 0 = 25 \]
\[ -\left(x\right)\left(x - 5\right) = 0 \]
\[ x = 0 \Rightarrow x = 0\qquad x - 5 = 0 \Rightarrow x = 5 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{25}}{-2} \]
\[ x = 0\quad\text{or}\quad x = 5 \]

The parabola y = -x² + 5x

xy
-0.5-2.75
0.52.25
1.55.25
2.56.25
3.55.25
4.52.25
5.5-2.75

Questions

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

The solutions of -x² + 5x = 0 are x = 0 or x = 5.

Does -x² + 5x = 0 factor?

Yes: -1(x)(x-5) = 0.

What is the vertex of y = -x² + 5x?

(5/2, 25/4). 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 = 0.

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