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 = 1/5 (25 + 3 i sqrt(5)) or x = (-3 i sqrt(5) + 25)/5

Step-by-step explanation using the quadratic formula:

Solve for x:

5 (x - 5)^2 + 9 = 0

Expand out terms of the left hand side:

5 x^2 - 50 x + 134 = 0

x = (50 ± sqrt((-50)^2 - 4×5×134))/(2×5) = (50 ± sqrt(2500 - 2680))/10 = (50 ± sqrt(-180))/10:

x = (50 + sqrt(-180))/10 or x = (50 - sqrt(-180))/10

sqrt(-180) = sqrt(-1) sqrt(180) = i sqrt(180):

x = (50 + i sqrt(180))/10 or x = (50 - i sqrt(180))/10

sqrt(180) = sqrt(4×9×5) = sqrt(2^2×3^2×5) = 2×3sqrt(5) = 6 sqrt(5):

x = (i×6 sqrt(5) + 50)/10 or x = (-i×6 sqrt(5) + 50)/10

Factor 2 from 50 + 6 i sqrt(5) giving 2 (3 i sqrt(5) + 25):

x = 1/102 (3 i sqrt(5) + 25) or x = (-6 i sqrt(5) + 50)/10

(2 (3 i sqrt(5) + 25))/10 = (2 (3 i sqrt(5) + 25))/(2×5) = (3 i sqrt(5) + 25)/5:

x = (3 i sqrt(5) + 25)/5 or x = (-6 i sqrt(5) + 50)/10

Factor 2 from 50 - 6 i sqrt(5) giving 2 (-3 i sqrt(5) + 25):

x = 1/5 (25 + 3 i sqrt(5)) or x = 1/102 (-3 i sqrt(5) + 25)

(2 (-3 i sqrt(5) + 25))/10 = (2 (-3 i sqrt(5) + 25))/(2×5) = (-3 i sqrt(5) + 25)/5:

Answer: x = 1/5 (25 + 3 i sqrt(5)) or x = (-3 i sqrt(5) + 25)/5

Answer 2

`(7\pm 3√(3)i)/(2)`

Explanation:

To solve :
`(x-5)^2 + 3(x -5) + 9 = 0`

On putting x-5 = u in the given equation, we get
`u^2+3u+9=0`

We will sove this equation using quadratic formula:

For equation of form
`ax^2+bx+c=0`, roots are given by
`x=(-b\pm √(b^2-4ac))/(2a)`

On comparing equation
`u^2+3u+9=0` with equation
`ax^2+bx+c=0`, we get
`a=1\,,\,b=3\,,\,c=9`

So, roots are
`u=(-3\pm √(9-36))/(2)=(-3\pm √(-27))/(2)=(-3\pm 3√(3)i)/(2)`

On putting u = x-5 in
`u=(-3\pm 3√(3)i)/(2)`, we get

`x-5=(-3\pm 3√(3)i)/(2)\Rightarrow x=5+\left ((-3\pm 3√(3)i)/(2)  \right )=(10-3\pm 3√(3)i)/(2)=(7\pm 3√(3)i)/(2)`