what is the solution of the equation $(x - 5)^2+3(x - 5)+9 = 0$? use $u$ substitution and the quadratic…

what is the solution of the equation $(x - 5)^2+3(x - 5)+9 = 0$? use $u$ substitution and the quadratic formula to solve.\n$x=\frac{-3pm3isqrt{3}}{2}$\n$x=\frac{7pm3isqrt{3}}{2}$\n$x = 2$\n$x = 8$

what is the solution of the equation $(x - 5)^2+3(x - 5)+9 = 0$? use $u$ substitution and the quadratic formula to solve.\n$x=\frac{-3pm3isqrt{3}}{2}$\n$x=\frac{7pm3isqrt{3}}{2}$\n$x = 2$\n$x = 8$

Answer

Answer:

A. $x=\frac{-3\pm3i\sqrt{3}}{2}$

Explanation:

Step1: Substitute $u = x - 5$

The equation $(x - 5)^2+3(x - 5)+9 = 0$ becomes $u^2+3u + 9=0$.

Step2: Recall quadratic formula

For a quadratic equation $au^2+bu + c = 0$ ($a = 1$, $b = 3$, $c = 9$), the solutions are given by $u=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

Step3: Calculate the discriminant

$\Delta=b^2 - 4ac=3^2-4\times1\times9=9 - 36=- 27$.

Step4: Find the value of $u$

$u=\frac{-3\pm\sqrt{-27}}{2}=\frac{-3\pm3i\sqrt{3}}{2}$.

Step5: Substitute back $u=x - 5$

$x-5=\frac{-3\pm3i\sqrt{3}}{2}$, so $x = 5+\frac{-3\pm3i\sqrt{3}}{2}=\frac{10-3\pm3i\sqrt{3}}{2}=\frac{7\pm3i\sqrt{3}}{2}$. There was a calculation - error in the above step. Let's correct it. $u=\frac{-3\pm\sqrt{-27}}{2}=\frac{-3\pm3i\sqrt{3}}{2}$, and since $u=x - 5$, then $x=5 + u=5+\frac{-3\pm3i\sqrt{3}}{2}=\frac{10-3\pm3i\sqrt{3}}{2}=\frac{7\pm3i\sqrt{3}}{2}$. The correct substitution and calculation: Let $u=x - 5$, the equation is $u^2+3u + 9 = 0$. Using the quadratic formula $u=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 1$, $b = 3$, $c = 9$. $b^{2}-4ac=3^{2}-4\times1\times9=9 - 36=-27$. $u=\frac{-3\pm\sqrt{-27}}{2}=\frac{-3\pm3i\sqrt{3}}{2}$. Since $u=x - 5$, then $x=5 + u$. $x=5+\frac{-3\pm3i\sqrt{3}}{2}=\frac{10-3\pm3i\sqrt{3}}{2}=\frac{7\pm3i\sqrt{3}}{2}$. The correct answer is $x=\frac{7\pm3i\sqrt{3}}{2}$.

Answer:

B. $x=\frac{7\pm3i\sqrt{3}}{2}$

Explanation:

Step1: Substitute $u=x - 5$

Let $u=x - 5$, then the equation $(x - 5)^2+3(x - 5)+9 = 0$ simplifies to $u^2+3u + 9 = 0$.

Step2: Apply quadratic formula

For a quadratic equation $au^2+bu + c = 0$ (here $a = 1$, $b = 3$, $c = 9$), the solutions are $u=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Calculate the discriminant $\Delta=b^2-4ac=3^2 - 4\times1\times9=9 - 36=-27$. Then $u=\frac{-3\pm\sqrt{-27}}{2}=\frac{-3\pm3i\sqrt{3}}{2}$.

Step3: Back - substitute $u$

Since $u=x - 5$, we solve for $x$: $x=u + 5=\frac{-3\pm3i\sqrt{3}}{2}+5=\frac{-3\pm3i\sqrt{3}+10}{2}=\frac{7\pm3i\sqrt{3}}{2}$.