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Solve 4x² + 9 = 0

4x² + 9 = 0 has no real solutions. The two complex solutions are x = (3i)/2 and x = (-3i)/2.

x² +x += 0
4x2+9=0⇒x=3i2, −3i2 (complex)4x^2 + 9 = 0 \Rightarrow x = \frac{3i}{2},\ \frac{-3i}{2}\ \text{(complex)}
Discriminant D = -144

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

Graph of y = 4x² + 9-4-3-2-1123410203040506070vertexy = 4x² + 9
Graph of y = 4x² + 9. The curve never reaches the x-axis, which is why there are no real solutions; the vertex is at (0, 9).

How to work it out

  1. 1

    Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 4, b = 0, c = 9.

    4x2+9=04x^2 + 9 = 0
  2. 2

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

    D=02−4⋅4⋅9=0−144=−144D = 0^2 - 4\cdot4\cdot9 = 0 - 144 = -144
  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=0±−1448x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-144}}{8}
  5. 5

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

    x=3i2orx=−3i2x = \frac{3i}{2}\quad\text{or}\quad x = \frac{-3i}{2}
LaTeX source of the formulas on this page
\[ 4x^2 + 9 = 0\ \Rightarrow\ x = \frac{3i}{2},\ \frac{-3i}{2} \]
\[ 4x^2 + 9 = 0 \]
\[ D = 0^2 - 4\cdot4\cdot9 = 0 - 144 = -144 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{0 \pm \sqrt{-144}}{8} \]
\[ x = \frac{3i}{2}\quad\text{or}\quad x = \frac{-3i}{2} \]

The parabola y = 4x² + 9

xy
-345
-225
-113
09
113
225
345

Questions

What are the solutions of 4x² + 9 = 0?

4x² + 9 = 0 has no real solutions. The two complex solutions are x = (3i)/2 and x = (-3i)/2.

Does 4x² + 9 = 0 factor?

Not over the integers, because the discriminant -144 is negative. Use the quadratic formula.

What is the vertex of y = 4x² + 9?

(0, 9). The parabola opens upwards, so this is the minimum.

What are the sum and product of the roots?

Sum = −b/a = 0, product = c/a = 9/4.

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