# How do you solve the quadratic with complex numbers given 10x^2-11x+9=13x-6x^2?

Mar 3, 2017

I think you want to solve for x, right?

$x = \frac{3}{4}$

There is only one solution.

#### Explanation:

$6 {x}^{2} + 10 {x}^{2} - 11 x - 13 x + 9 = 0$

$16 {x}^{2} - 24 x + 9 = 0$

$x = \frac{- \left(- 24\right) \pm \sqrt{{\left(- 24\right)}^{2} - 4 \left(9 \cdot 16\right)}}{2 \cdot 16}$

$x = \frac{24 \pm \sqrt{\left(576 - 4 \left(144\right)\right)}}{32}$

$x = \frac{24 \pm \sqrt{\left(576 - 576\right)}}{32}$

$x = \frac{24}{32} = \frac{3}{4}$

Since under the square root is equal to 0, this means that there is only one solution.

From looking at the graph, we can also see that it is equal to $\frac{3}{4}$
graph{16x^2-24x+9 [-1.276, 2.436, -0.928, 0.928]}

Mar 6, 2017

$x = \frac{3}{4}$

#### Explanation:

$10 {x}^{2} - 11 x + 9 = 13 x - 6 {x}^{2} \text{ } \leftarrow$ make a quadratic equal to 0

$10 {x}^{2} + 6 {x}^{2} - 11 x - 13 x + 9 = 0$

$16 {x}^{2} - 24 x + 9 = 0 \text{ } \leftarrow$ find factors

(Find factors of 16 and 9 which add to give 24.)

$\left(4 x - 3\right) \left(4 x - 3\right) = 0$

$4 x - 3 = 0$

$4 x = 3$

$x = \frac{3}{4}$

As the two brackets are the same, there is only one solution.