3x^2 + 2x - k = 0, find 3α^2 - 2β in terms of k

[tony24810]

Homework Statement

Let k be a constant. If α and β are the roots of the equation 3x^2 + 2x - k = 0, find the value of 3α^2 - 2β in terms of k.

Homework Equations

The Attempt at a Solution

Obviously, the usual

αβ = -k/3
α + β = -2/3

has been written but I couldn't put them into the equation required despite a full hour's effort.

Also, tried writing (-b+-sqrt(b^2-4ac))/2a, and put respective roots into the equation, it yields something similar to the provided solution, but has an extra root term.

The solution is 4/3 + k.

The latter method gets 4/3 + k + sqrt(4+12k)/6.

 

[Mark44]

Homework Statement

Let k be a constant. If α and β are the roots of the equation 3x^2 + 2x - k = 0, find the value of 3α^2 - 2β in terms of k.

Homework Equations

The Attempt at a Solution

Obviously, the usual

αβ = -k/3
α + β = -2/3

has been written but I couldn't put them into the equation required despite a full hour's effort.

Also, tried writing (-b+-sqrt(b^2-4ac))/2a, and put respective roots into the equation, it yields something similar to the provided solution, but has an extra root term.

The solution is 4/3 + k.

The latter method gets 4/3 + k + sqrt(4+12k)/6.

Considering k to be a constant, solve the second equation below for α or β, then substitute into the first equation.
αβ = -k/3
α + β = -2/3

 

[tony24810]

Considering k to be a constant, solve the second equation below for α or β, then substitute into the first equation.
αβ = -k/3
α + β = -2/3

It has a β^2 term leftover.

 

[ehild]

Homework Statement

Let k be a constant. If α and β are the roots of the equation 3x^2 + 2x - k = 0, find the value of 3α^2 - 2β in terms of k.

Homework Equations

The Attempt at a Solution

Obviously, the usual

αβ = -k/3
α + β = -2/3

has been written but I couldn't put them into the equation required despite a full hour's effort.

Also, tried writing (-b+-sqrt(b^2-4ac))/2a, and put respective roots into the equation, it yields something similar to the provided solution, but has an extra root term.

The solution is 4/3 + k.

The latter method gets 4/3 + k + sqrt(4+12k)/6.

Show your work in detail.

ehild

 

[tony24810]

haha omg i got it

 

[tony24810]

got it

i tried again this time giving each equation their names, suddenly spot the identity that i didn't see before. hahahaha