Respuesta :
Answer:
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:
[tex]\frac{7\pm 3\sqrt{3}i}{2}[/tex]
Step-by-step explanation:
To solve : [tex](x-5)^2 + 3(x -5) + 9 = 0[/tex]
On putting x-5 = u in the given equation, we get [tex]u^2+3u+9=0[/tex]
We will sove this equation using quadratic formula:
For equation of form [tex]ax^2+bx+c=0[/tex], roots are given by [tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
On comparing equation [tex]u^2+3u+9=0[/tex] with equation [tex]ax^2+bx+c=0[/tex], we get [tex]a=1\,,\,b=3\,,\,c=9[/tex]
So, roots are [tex]u=\frac{-3\pm \sqrt{9-36}}{2}=\frac{-3\pm \sqrt{-27}}{2}=\frac{-3\pm 3\sqrt{3}i}{2}[/tex]
On putting u = x-5 in [tex]u=\frac{-3\pm 3\sqrt{3}i}{2}[/tex], we get
[tex]x-5=\frac{-3\pm 3\sqrt{3}i}{2}\Rightarrow x=5+\left (\frac{-3\pm 3\sqrt{3}i}{2} \right )=\frac{10-3\pm 3\sqrt{3}i}{2}=\frac{7\pm 3\sqrt{3}i}{2}[/tex]
