Question

What is the solution of the equation (x – 5)2 + 3(x – 5) + 9 = 0? Use u substitution and the quadratic formula to solve.

Answer 1

(x – 5)² + 3(x – 5) + 9 = 0

x² - 10x + 25 + 3x - 15 + 9 = 0

x² - 7x - 16 = 0

$x = \frac{7 + \sqrt{(-7)^2-4(1)(-16)} }{2}$ or $x = \frac{7 - \sqrt{(-7)^2-4(1)(-16)} }{2}$ 

$x = \frac{7 + 3i \sqrt{3} }{2}$ or $x = \frac{7 - 3i \sqrt{3} }{2}$ 

(Answer B)

Answer 2

Answer: $x=\frac{7\pm 3i\sqrt{3}}{2}$

Explanation:

Since, given quadratic function, $(x-5)^2 + 3(x-5) + 9 = 0$ -----(1)

let us consider,$x-5=p$ -----(2)

Put this value in equation (1),

We get, $p^2+3 p+9=0$, which is also a quadratic equation.

Since, quadratic formula for finding the root of quadratic equation of type $ax^2+bx+c=0$ is,  $x=\frac{-b\pm\sqrt{b^2-4ac}} {2a}$

Here, a=1, b=3  and c=9, so, $p=\frac{-3\pm\sqrt{3^2-4\times1 \times9}} {2\times1}$

Thus,  $p=\frac{-3\pm3\sqrt{-3}} {2}$ ⇒ $p=\frac{-3\pm3i\sqrt{3}} {2}$   (because √-1=i)

Now, from equation(2) $x-5=\frac{-3\pm3i\sqrt{3}} {2}$

⇒$x=\frac{-3\pm3i\sqrt{3}} {2}+5=\frac{7\pm 3i\sqrt{3}}{2}$

⇒$x=\frac{7\pm 3i\sqrt{3}}{2}$