Verified answer

Since A and B are roots of 2x^2 - 3x + 4 = 0, then x - A and x - B are factors of 2x^2 - 3x + 4 by the Factor Theorem. From this, we can write:

2x^2 - 3x + 4 = 2(x - A)(x - B).

Note that the factor of 2 out front is because the x^2 coefficient of (x - A)(x - B) is 1 and the extra factor of 2 brings it up to 2 as needed.

By FOILing the right side:

2(x - A)(x - B) = 2(x^2 - Ax - Bx + AB) = 2x^2 - [2(A + B)]x + 2AB.

So:

2x^2 - 3x + 4 = 2x^2 - [2(A + B)]x + 2AB.

These two polynomials are the same, so their coefficients all must be equal. This gives:

2(A + B) = 3 and 2AB = 4 ==> A + B = 3/2 and 2AB = 4.

Therefore, since (A + B)^2 = A^2 + 2AB + B^2:

(3/2)^2 = A^2 + B^2 + 4 ==> A^2 + B^2 = -7/4.

I hope this helps!

A+B = 3/2 and AB = 4/2 = 2

=> A^2 + B^2

= (A+B)^2 - 2AB

= (3/2)^2 - 2*2

= 9/4 - 4

= - 7/4

[Note that the sum of squares is negative indicates that the roots A and B are complex.]

A + B = 3/2

AB = 2

and (A + B)² = A² + B² + 2AB

i.e. A² + B² = (A + B)² - 2AB

=> A² + B² = (3/2)² - 2(2) = 9/4 - 4

so, A² + B² = -7/4

:)>