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Solve x² - 6x + 5 = 0

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

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

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

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Graph of y = x² - 6x + 5-11234567-551015x = 5x = 1vertexy = x² - 6x + 5
Graph of y = x² - 6x + 5. The curve crosses the x-axis at x = 5 and x = 1; 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=0x^2 - 6x + 5 = 0
  2. 2

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

    D=(−6)2−4⋅1⋅5=36−20=16D = (-6)^2 - 4\cdot1\cdot5 = 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−5)(x−1)=0\left(x - 5\right)\left(x - 1\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−5=0⇒x=5x−1=0⇒x=1x - 5 = 0 \Rightarrow x = 5\qquad x - 1 = 0 \Rightarrow x = 1
  6. 6

    Method 2, The quadratic formula works for every quadratic:

    x=−b±D2a=6±162x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{16}}{2}
  7. 7

    Simplify to get the solutions:

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

    Check by putting x = 5 back in: a·x² + b·x + c = 1·(25) + (-6)·(5) + 5 = 0 ✓

LaTeX source of the formulas on this page
\[ x^2 - 6x + 5 = 0\ \Rightarrow\ x = 5,\ 1 \]
\[ x^2 - 6x + 5 = 0 \]
\[ D = (-6)^2 - 4\cdot1\cdot5 = 36 - 20 = 16 \]
\[ \left(x - 5\right)\left(x - 1\right) = 0 \]
\[ x - 5 = 0 \Rightarrow x = 5\qquad x - 1 = 0 \Rightarrow x = 1 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{16}}{2} \]
\[ x = 5\quad\text{or}\quad x = 1 \]

The parabola y = x² - 6x + 5

xy
05
10
2-3
3-4
4-3
50
65

Questions

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

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

Does x² - 6x + 5 = 0 factor?

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

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

(3, -4). 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 = 5.

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