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Solve 5x² - 7x + 4 = 0

5x² - 7x + 4 = 0 has no real solutions. The two complex solutions are x = (7 + √31 i)/10 and x = (7 - √31 i)/10.

x² +x += 0
5x2−7x+4=0⇒x=7+31 i10, 7−31 i10 (complex)5x^2 - 7x + 4 = 0 \Rightarrow x = \frac{7 + \sqrt{31}\, i}{10},\ \frac{7 - \sqrt{31}\, i}{10}\ \text{(complex)}
Discriminant D = -31

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

Graph of y = 5x² - 7x + 4-3-2-1123420406080vertexy = 5x² - 7x + 4
Graph of y = 5x² - 7x + 4. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0.7, 1.55).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 5, b = -7, c = 4.

    5x2−7x+4=05x^2 - 7x + 4 = 0
  2. 2

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

    D=(−7)2−4⋅5⋅4=49−80=−31D = (-7)^2 - 4\cdot5\cdot4 = 49 - 80 = -31
  3. 3

    D is negative, so there are no real solutions; the two solutions are complex numbers.

  4. 4

    The quadratic formula works for every quadratic:

    x=−b±D2a=7±−3110x = \frac{-b \pm \sqrt{D}}{2a} = \frac{7 \pm \sqrt{-31}}{10}
  5. 5

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

    x=7+31 i10orx=7−31 i10x = \frac{7 + \sqrt{31}\, i}{10}\quad\text{or}\quad x = \frac{7 - \sqrt{31}\, i}{10}
LaTeX source of the formulas on this page
\[ 5x^2 - 7x + 4 = 0\ \Rightarrow\ x = \frac{7 + \sqrt{31}\, i}{10},\ \frac{7 - \sqrt{31}\, i}{10} \]
\[ 5x^2 - 7x + 4 = 0 \]
\[ D = (-7)^2 - 4\cdot5\cdot4 = 49 - 80 = -31 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{7 \pm \sqrt{-31}}{10} \]
\[ x = \frac{7 + \sqrt{31}\, i}{10}\quad\text{or}\quad x = \frac{7 - \sqrt{31}\, i}{10} \]

The parabola y = 5x² - 7x + 4

xy
-2.346.55
-1.321.55
-0.36.55
0.71.55
1.76.55
2.721.55
3.746.55

Questions

What are the solutions of 5x² - 7x + 4 = 0?

5x² - 7x + 4 = 0 has no real solutions. The two complex solutions are x = (7 + √31 i)/10 and x = (7 - √31 i)/10.

Does 5x² - 7x + 4 = 0 factor?

Not over the integers, because the discriminant -31 is negative. Use the quadratic formula.

What is the vertex of y = 5x² - 7x + 4?

(7/10, 31/20). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = 7/5, product = c/a = 4/5.

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