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Factorize (x2 -0.7x+0.49) Using imaginary numbers, into two terms such that:  (x- something) (x- something) ?

Guest Apr 30, 2017
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 #1
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Solve for x:
x^2 - 0.7 x + 0.49 = 0

x^2 - 0.7 x + 0.49 = x^2 - (7 x)/10 + 49/100:
x^2 - (7 x)/10 + 49/100 = 0

Subtract 49/100 from both sides:
x^2 - (7 x)/10 = -49/100

Add 49/400 to both sides:
x^2 - (7 x)/10 + 49/400 = -147/400

Write the left hand side as a square:
(x - 7/20)^2 = -147/400

Take the square root of both sides:
x - 7/20 = (7 i sqrt(3))/20 or x - 7/20 = -(7 i sqrt(3))/20

Add 7/20 to both sides:
x = 7/20 + (7 i sqrt(3))/20 or x - 7/20 = -(7 i sqrt(3))/20

Add 7/20 to both sides:
Answer: | x = 7/20 + (7 i sqrt(3))/20       or        x = 7/20 - (7 i sqrt(3))/20

Guest Apr 30, 2017
 #2
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Factorize (x2 -0.7x+0.49) Using imaginary numbers, into two terms such that:  (x- something) (x- something) ?

 

\(If\;\;x^2 -0.7x+0.49=0\;\;then\\x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {--0.7 \pm \sqrt{0.49-4*0.49} \over 2}\\ x = {0.7 \pm \sqrt{-3*0.49} \over 2}\\ x = {0.7 \pm 0.7 \sqrt{3}i \over 2}\\ x = 0.35 \pm 0.35 \sqrt{3}i \\ x = 0.35(1+\sqrt{3}i )\quad or\quad x = 0.35(1-\sqrt{3}i )\\\\~\\ so\\ x^2 -0.7x+0.49=(x-0.35(1+\sqrt{3}i )(x+0.35(1+\sqrt{3}i ))\\ x^2 -0.7x+0.49=(x-0.35-0.35\sqrt{3}i )(x+0.35+0.35\sqrt{3}i )\)

Melody  Apr 30, 2017
 #3
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It would have been to both our benefits if you had mentioned yesterday that you were looking for real or imaginary factors.

This is a young persons forum so it is inappropriate for me to assume any question asker knows anything about complex numbers!

Melody  Apr 30, 2017

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