Respuesta :

[tex]B)\frac{2-i\sqrt[]{5\text{ }}}{2}[/tex]

Explanation

[tex]4x^2=8x-9[/tex]

Step 1

set the left side equal to zero

[tex]\begin{gathered} 4x^2=8x-9 \\ \text{subtract }4x^2\text{ in boht sides} \\ 4x^2-4x^2=8x-9=8x-9-4x^2=8x-9 \\ 0=-4x^2+8x-9 \\ \end{gathered}[/tex]

Step 2

solve for x,

use the quadratic formula:

it says

[tex]\begin{gathered} \text{for} \\ ax^2+bx+c=0 \\ \text{the solution for x is} \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]

then

[tex]\begin{gathered} ax^2+bx+c=0\Rightarrow-4x^2+8x-9 \\ so \\ a=-4 \\ b=8 \\ c=-9 \end{gathered}[/tex]

replace and calculate

[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-8\pm\sqrt[]{8^2-4(-4)(-9)}}{2(-4)} \\ x=\frac{-8\pm\sqrt[]{64-144}}{-8} \\ x=\frac{-8\pm\sqrt[]{-80}}{-8} \\ x=\frac{-8\pm\sqrt[]{-16\cdot5}}{-8} \\ x=\frac{-8\pm4\sqrt[]{-\cdot5}}{-8}=\frac{4(-2\pm i\sqrt[]{5})}{4(-2)} \\ x=\frac{-2\pm i\sqrt[]{5}}{-2} \end{gathered}[/tex]

therefore, the solutions are

[tex]\begin{gathered} x=\frac{-2\pm i\sqrt[]{5}}{-2} \\ x_2=\frac{-2+i\sqrt[]{5}}{-2}=\frac{-(2-i\sqrt[]{5\text{ )}}}{-2}=\frac{2-i\sqrt[]{5\text{ }}}{2} \\ x_2=\frac{-2-i\sqrt[]{5}}{-2}=\frac{-(2+i\sqrt[]{5\text{ )}}}{-2}=\frac{2+i\sqrt[]{5\text{ }}}{2} \end{gathered}[/tex]

therefore, the answer is

[tex]B)\frac{2-i\sqrt[]{5\text{ }}}{2}[/tex]

I hope this helps you

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