The roots of the equation 3x²-2x-4=0 are j and k. evaluate j² + k².

28/9

Step-by-step explanation:

If the roots are J and K, then:

3 (x − J) (x − K) = 0

3 (x² − (J+K)x + JK) = 0

So if we factor out the leading coefficient:

3x² − 2x − 4 = 0

3(x² − 2/3x − 4/3) = 0

The coefficient of the second term is the sum of the roots:

J + K = 2/3

And the constant is the product of the roots:

JK = -4/3

If we take the sum of the roots and square it:

(J + K)² = (2/3)²

J² + 2JK + K² = 4/9

And subtract twice the product:

J² + K² = 4/9 − 2JK

J² + K² = 4/9 − 2(-4/3)

J² + K² = 4/9 + 8/3

J² + k² = 28/9

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

25.45584