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1 day ago

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

Mathematics

2 answers:

PIT_PIT [3.9K]1 day ago

8 0

Answer:

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

Detailed solution using the quadratic formula:

We begin by solving for x:

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

Expand the left side:

5 x^2 - 50 x + 134 = 0

Using the quadratic formula:

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

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

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

For sqrt(180), we compute: sqrt(4×9×5) = sqrt(2^2×3^2×5) = 2×3sqrt(5) = 6 sqrt(5):

x can be expressed as (i×6 sqrt(5) + 50)/10 or (-i×6 sqrt(5) + 50)/10

Factoring out 2 from 50 + 6 i sqrt(5) gives us 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 simplifies to (2 (3 i sqrt(5) + 25))/(2×5) = (3 i sqrt(5) + 25)/5:

x can be either (3 i sqrt(5) + 25)/5 or (-6 i sqrt(5) + 50)/10

Factoring out 2 from 50 - 6 i sqrt(5) yields 2 (-3 i sqrt(5) + 25):

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

(2 (-3 i sqrt(5) + 25))/10 can be simplified to (2 (-3 i sqrt(5) + 25))/(2×5) = (-3 i sqrt(5) + 25)/5:

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

AnnZ [3.8K]1 day ago

6 0

Answer:

$\frac{7\pm 3\sqrt{3}i}{2}$

Step-by-step explanation:

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

Substituting x-5 = u into the original equation provides us with $u^2+3u+9=0$

We will address this equation through the quadratic formula:

In the equation of form $ax^2+bx+c=0$ , the roots are determined by $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Comparing equation $u^2+3u+9=0$ with $ax^2+bx+c=0$ , we find $a=1\,,\,b=3\,,\,c=9$

Thus, the roots are $u=\frac{-3\pm \sqrt{9-36}}{2}=\frac{-3\pm \sqrt{-27}}{2}=\frac{-3\pm 3\sqrt{3}i}{2}$

Transforming u back to x in $u=\frac{-3\pm 3\sqrt{3}i}{2}$ , leads us to

$x-5=\frac{-3\pm 3\sqrt{3}i}{2}\Rightarrow x=5+\left (\frac{-3\pm 3\sqrt{3}i}{2} \right )=\frac{10-3\pm 3\sqrt{3}i}{2}=\frac{7\pm 3\sqrt{3}i}{2}$

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