# let a and b be the solution of 3x^2-x-3=0.Then find a^2+b^2

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### 3 Answers

We can do this much easier. Sorry for my previous answer.

So we can factor to find

c(x - a)(x - b) = 0

cx^2 - c(a+b)x + abc = 0

since this must be the same as 3x^2-x-3 = 0

so c = 3, a+b = -1/3 and ab = -1

since (a+b) = -1/3 then (a+b)^2 = 1/9 and we can simplify

a^2 + 2ab + b^2 = 1/9 and we know that ab = -1 so

a^2 + 2(-1) + b^2 = 1/9 simplifying we get

**a^2 + b^2 = 1/9 + 2 = 19/9 **

So we can factor to find

c(x - a)(x - b) = 0

cx^2 - c(a+b)x + abc = 0

since this must be the same as 3x^2-x-3 = 0

so c = 3, a+b = -1/3 and ab = -1

since (a+b) = -1/3 then (a+b)^2 = 1/9 and we can simplify

a^2 + 2ab + b^2 = 1/9 and we know that ab = -1 so

a^2 + 2(-1) + b^2 = 1/9 simplifying we get

a^2 + b^2 = 1/9 + 2 = 19/9

since 3x^2 - x - 3 is not factorable, we will have to solve the equation 3x^2 - x - 3 = 0 using the quadratic formula.

the quadratic formula is:

let ax^2 + bx + c = 0

x = [-b + sqrt(b^2 - 4ac)]/2a or x = [-b - sqrt(b^2 - 4ac)]/2a

with the equation 3x^2 - x - 3 = 0, we get:

a = 3, b = -1, c = -3

using this, substitution gives us the following expressions for x:

x(1) = [-(-1)+sqrt((-1)^2 - 4(3)(-3))]/2(3)

= [1+sqrt(1 + 36)]/6

= [1 + sqrt(37)]/6

x(2) = [-(-1)-sqrt((-1)^2 - 4(3)(-3))]/2(3)

= [1-sqrt(1 + 36)]/6

= [1 - sqrt(37)]/6

since x(1) and x(2) are the solutions to the equation, let a = x(1) and b = x(2).

a = [1+sqrt(37)]/6

a^2 = {[1+sqrt(37)]/6}^2

= [1+sqrt(37)]^2/6^2

= [1+2sqrt(37)+sqrt(37)^2]/36

= [1+2sqrt(37)+37]/36

= [38+2sqrt(37)]/36

= [19+sqrt(37)]/18

b = [1-sqrt(37)]/6

b^2 = {[1-sqrt(37)]/6}^2

= [1-sqrt(37)]^2/6^2

= [1-2sqrt(37)+sqrt(37)^2]/36

= [1-2sqrt(37)+37]/36

= [38-2sqrt(37)]/36

= [19-sqrt(37)]/18

a^2 + b^2 = [19+sqrt(37)]/18 + [19-sqrt(37)]/18

= [19+sqrt(37)+19-sqrt(37)]/18

= [19 + 19]/18

= 38/18

= 19/9 or 2 1/9