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avatar+1679 

Find the sum of the squares of the roots of $2x^2+4x-1=x^2-8x+3$.

 Jan 11, 2024
 #1
avatar+289 
+1

\(2x^2+4x-1=x^2-8x+3\)

 

We can make this is the form  

 

\(ax^2+bx+c=0\)

 

We get:

 

\(x^2+12x-4=0\)

 

a = 1, b = 12, and c = -4.

 

Using the quadratic formula \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\), we get:

 

\(x = {-12 \pm \sqrt{12^2+16} \over 2}\)

 

\(x = {-12 \pm \sqrt{160} \over 2}\)

 

\(x = {-12 - \sqrt{160} \over 2}\)\(x = {-12 + \sqrt{160} \over 2}\)

 

Now we can square both values of x and then add them:

 

-151.894663844 + 0.10533615595 = -151.789327688

 

Answer: -151.789327688

 Jan 11, 2024
 #2
avatar+36919 
+1

Re-arrange to 

x^2 +12x -4 = 0     complete the square 

(x+6)^2  = 4 + 36 

x+6 = +- sqrt 40 

x =  - 6 +- sqrt 40      <==== these are the roots 

 

Now square the roots

(-6 + sqrt 40)2 = 76 - 12 sqrt 40 

(-6  - sqrt 40)= 76 +12 sqrt 40

                        added together =  76 + 76 =   + 152  

 Jan 11, 2024
 #3
avatar+128732 
+1

Simplify as

 

x^2 + 12x - 4 =  0

 

Call the roots a,b

 

Product of  roots

ab  = -4

2ab  =  -8

 

Sum of  roots 

a + b  =  -12      square both sides

a^2 + 2ab + b^2 = 144

a^2 + b^2  -8   =144

a^2 + b^2   = 152

 

cool cool cool

 Jan 12, 2024
 #4
avatar+53 
0

PLug in the quadratic formula:\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 Jan 12, 2024

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