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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, −5x^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 + 5-7-6-5-4-3-2-11-551015x = -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=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=62−4⋅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+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±162x = \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
\[ x^2 + 6x + 5 = 0\ \Rightarrow\ x = -1,\ -5 \]
\[ x^2 + 6x + 5 = 0 \]
\[ D = 6^2 - 4\cdot1\cdot5 = 36 - 20 = 16 \]
\[ \left(x + 1\right)\left(x + 5\right) = 0 \]
\[ x + 1 = 0 \Rightarrow x = -1\qquad x + 5 = 0 \Rightarrow x = -5 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{16}}{2} \]
\[ x = -1\quad\text{or}\quad x = -5 \]

The parabola y = x² + 6x + 5

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

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: (x+1)(x+5) = 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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