reciprocals of the solutions

dude are you sure about question some value is wrong

On simplification we get 15x^2 - 9x + 55 = 0 (bring the -11/3 to LHS and multiply throughout by 15).

Now we know if a and b are the roots of a quadratic, a + b = -middle term constant / first term constant and ab = last term constant / first term, thus (a+b)/ab = 1/a + 1/b = -middle term/ last term

=-(-9)/55 = 9/55

B

OA please?

amising6 wrote:dude are you sure about question some value is wrong

such as?

mj78ind wrote:On simplification we get 15x^2 - 9x + 55 = 0 (bring the -11/3 to LHS and multiply throughout by 15).

Now we know if a and b are the roots of a quadratic, a + b = -middle term constant / first term constant and ab = last term constant / first term, thus (a+b)/ab = 1/a + 1/b = -middle term/ last term

=-(-9)/55 = 9/55

B

OA please?

A quadratic in the form x^2 – S x + P = 0 has S as the sum and P as the product of its two solutions. That is if α and β were the two solutions, then the sum of their reciprocals will be (α + β) / (α β) = S/P. [spoiler]You got it right![/spoiler]