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Solve 2x² + 11x - 5 = 0

The solutions of 2x² + 11x - 5 = 0 are x = (-11 + √161)/4 or x = (-11 - √161)/4, that is x ≈ 0.42214439 and x ≈ -5.9221444.

x² +x += 0
2x2+11x−5=0⇒x=−11+1614, −11−16142x^2 + 11x - 5 = 0 \Rightarrow x = \frac{-11 + \sqrt{161}}{4},\ \frac{-11 - \sqrt{161}}{4}
Discriminant D = 161 · x ≈ 0.422144, -5.92214

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

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Graph of y = 2x² + 11x - 5-8-6-4-22-20-1010203040x = 0.42214439x = -5.9221444vertexy = 2x² + 11x - 5
Graph of y = 2x² + 11x - 5. The curve crosses the x-axis at x = 0.42214439 and x = -5.9221444; the vertex is at (-2.75, -20.125).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 2, b = 11, c = -5.

    2x2+11x−5=02x^2 + 11x - 5 = 0
  2. 2

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

    D=112−4⋅2⋅(−5)=121+40=161D = 11^2 - 4\cdot2\cdot(-5) = 121 + 40 = 161
  3. 3

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

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=−11±1614x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-11 \pm \sqrt{161}}{4}
  5. 5

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

    x=−11+1614orx=−11−1614x = \frac{-11 + \sqrt{161}}{4}\quad\text{or}\quad x = \frac{-11 - \sqrt{161}}{4}
  6. 6

    As decimals: x ≈ 0.42214439 and x ≈ -5.9221444.

  7. 7

    Check numerically: for x ≈ 0.42214439 the left side is -8.8817842e-16 (zero up to rounding) ✓

LaTeX source of the formulas on this page
\[ 2x^2 + 11x - 5 = 0\ \Rightarrow\ x = \frac{-11 + \sqrt{161}}{4},\ \frac{-11 - \sqrt{161}}{4} \]
\[ 2x^2 + 11x - 5 = 0 \]
\[ D = 11^2 - 4\cdot2\cdot(-5) = 121 + 40 = 161 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-11 \pm \sqrt{161}}{4} \]
\[ x = \frac{-11 + \sqrt{161}}{4}\quad\text{or}\quad x = \frac{-11 - \sqrt{161}}{4} \]

The parabola y = 2x² + 11x - 5

xy
-5.75-2.125
-4.75-12.125
-3.75-18.125
-2.75-20.125
-1.75-18.125
-0.75-12.125
0.25-2.125

Questions

What are the solutions of 2x² + 11x - 5 = 0?

The solutions of 2x² + 11x - 5 = 0 are x = (-11 + √161)/4 or x = (-11 - √161)/4, that is x ≈ 0.42214439 and x ≈ -5.9221444.

Does 2x² + 11x - 5 = 0 factor?

Not over the integers, because the discriminant 161 is not a perfect square. Use the quadratic formula.

What is the vertex of y = 2x² + 11x - 5?

(-11/4, -161/8). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = -11/2, product = c/a = -5/2.

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    Solve 2x² + 11x - 5 = 0: x = (-11 + √161)/4 or x = (-11 - √161)/4 (steps) | WordGPT