Question 61109This question is from textbook college algebra
: Solve the following equations by completing the square.
x^2+8x+7=-8
2x^2=7x-15=0
u^2+10u+9=0
Solve the quadratic by using the quadratic formula.
4x^2-3x=-1
This question is from textbook college algebra
Answer by pvaka(23)   (Show Source):
You can put this solution on YOUR website! This is four differnt quetions...
Solve the following equations by completing the square.
(1.) x^2+8x+7=-8
=x^2+8x+7+8= -8+8
=x^2+8x+15=0
=x^2+3x+5x+15=0
=(x^2+3x)+(5x+15)=0
=x(x+3)+5(x+3)
=(x+5)(x+3)=0
>
x+5=0,
x=-5
x+3=0
x=-3
x={-5,-3} solution
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(2.) 2x^2=7x-15=0
=2x^2-10x+3x-15=0
=(2x^2-10x)+(3x-15)
=2x(x-5)+3(x-5)
=(2x+3)(x-5)=0
>>
=2x+3=0
=x=-3/2
x-5=0
x=5
x={-3/2,5} solution
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(3.)u^2+10u+9=0
=u^2+ u+9u+9=0
=(u^2+u)+(9u+9)=0
=u(u+1)+9(u+1)=0
=(u+9)(u+1)=0
>>
=u+1=0
=u=-1
u+9=0
u=-9
x={-9,-1} solution
---------------------------------------------------------------
Solve the quadratic by using the quadratic formula.
4x^2-3x=-1
=4x^2-3x+1=-1+1
=4x^2-3x+1=0----> a=4,b=-3,c=1
>>
x=(-(-3)ħSQRT((-3)^2-4*4*1))/2*4
x=(3ħSQRT(9-16))/8
x=(3ħSQRT(-7))/8
>>
If you are working with real numbers only, this is as far as you can go, since it is not possible to find the square root of a negative number in the real number system.
So the answer would be :"There are no real solutions"
>>
However, if you are working in the complex number system, you can keep going :
>
x = (3 ħ i SQRT 7)/8
>
So in this case, there are two complex conjugate solutions :
x=( 3 ħ i SQRT 7)/8
These solutions are approximately
x = 0.375 + 0.330719 i , x = 0.375 - 0.330719 i
Pvaka
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