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# Algebra

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24
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+1678

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

Jan 11, 2024

#1
+290
+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

Jan 11, 2024
#2
+37045
+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
+129829
+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

Jan 12, 2024
#4
+57
0

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

Jan 12, 2024