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Solve -x² + 6x - 5 = 0

The solutions of -x² + 6x - 5 = 0 are x = 1 or x = 5.

x² +x += 0
−x2+6x−5=0⇒x=1, 5-x^2 + 6x - 5 = 0 \Rightarrow x = 1,\ 5
Discriminant D = 16 · x ≈ 1, 5

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

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Graph of y = -x² + 6x - 5-11234567-15-10-55x = 1x = 5vertexy = -x² + 6x - 5
Graph of y = -x² + 6x - 5. The curve crosses the x-axis at x = 1 and x = 5; the vertex is at (3, 4).

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 = -5.

    −x2+6x−5=0-x^2 + 6x - 5 = 0
  2. 2

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

    D=62−4⋅(−1)⋅(−5)=36−20=16D = 6^2 - 4\cdot(-1)\cdot(-5) = 36 - 20 = 16
  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−1)(x−5)=0-\left(x - 1\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−1=0⇒x=1x−5=0⇒x=5x - 1 = 0 \Rightarrow x = 1\qquad x - 5 = 0 \Rightarrow x = 5
  6. 6

    Method 2, The quadratic formula works for every quadratic:

    x=−b±D2a=−6±16−2x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{16}}{-2}
  7. 7

    Simplify to get the solutions:

    x=1orx=5x = 1\quad\text{or}\quad x = 5
  8. 8

    Check by putting x = 1 back in: a·x² + b·x + c = -1·(1) + 6·(1) + (-5) = 0 ✓

LaTeX source of the formulas on this page
\[ -x^2 + 6x - 5 = 0\ \Rightarrow\ x = 1,\ 5 \]
\[ -x^2 + 6x - 5 = 0 \]
\[ D = 6^2 - 4\cdot(-1)\cdot(-5) = 36 - 20 = 16 \]
\[ -\left(x - 1\right)\left(x - 5\right) = 0 \]
\[ x - 1 = 0 \Rightarrow x = 1\qquad x - 5 = 0 \Rightarrow x = 5 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{16}}{-2} \]
\[ x = 1\quad\text{or}\quad x = 5 \]

The parabola y = -x² + 6x - 5

xy
0-5
10
23
34
43
50
6-5

Questions

What are the solutions of -x² + 6x - 5 = 0?

The solutions of -x² + 6x - 5 = 0 are x = 1 or x = 5.

Does -x² + 6x - 5 = 0 factor?

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

What is the vertex of y = -x² + 6x - 5?

(3, 4). 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 = 5.

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