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Paha777 [63]
3 years ago
7

Solve the equation by completing the square. x^2-7x-4=0. Have to show steps.

Mathematics
1 answer:
Paul [167]3 years ago
4 0
X^2-7x-4=0
(x-7/2)^2-(7/2)^2-4=0
(x-7/2)^2-(7)^2/(2)^2-4=0
(x-7/2)^2-49/4-4=0
(x-7/2)^2+(-49-4*4)/4=0
(x-7/2)^2+(-49-16)/4=0
(x-7/2)^2+(-65)/4=0
(x-7/2)^2-65/4=0
(x-7/2)^2-65/4+65/4=0+65/4
(x-7/2)^2=65/4
sqrt[ (x-7/2)^2 ]=+-sqrt(65/4)
x-7/2=+-sqrt(65)/sqrt(4)
x-7/2=+-sqrt(65)/2
x-7/2+7/2=+-sqrt(65)/2+7/2
x=7/2+-sqrt(65)/2
x=[7+-sqrt(65)]/2
x1=[7-sqrt(65)]/2
x2=[7+sqrt(65)]/2
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Answer:

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Step-by-step explanation:

p(x) = 4x - 3

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Step-by-step explanation:

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Step-by-step explanation:

Given:

f(x)=x^3+7x^2+7x-15

Finding all the possible rational zeros of f(x)

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q=±1(factors of coefficient of leading term)

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Now finding the rational zeros using rational root theorem

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f(1)=1+7+7-15

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f(-1)= -1 +7-7-15

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f(3)=27+7(9)+21-15

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f(-3)= (-3)^3+7(-3)^2+7(-3)-15

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f(5)=5^3+7(5)^2+7(5)-15

    =320

f(-5)=(-5)^3+7(-5)^2+7(-5)-15

      =0    

f(15)=(15)^3+7(15)^2+7(15)-15

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f(-15)=(-15)^3+7(-15)^2+7(-15)-15

      =-1920

Hence the rational roots are 1,-3,-5 !

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