Question

If the sum of the zeros of the quadratic polynomial kx×x+2x+3k is equal to the product of its zeros , then find the value of 'k'​

Answer 1

Answer: -2/3

Step-by-step explanation:

p(x)=kx^2+2x+3k

f(x)= {-2+√(4-12k^2)}÷2k ......f(x1)

= {-2-√(4-12k^2)}÷2k ......f(x2)

A/q,

f(x1)+f(x2)=f(x1.x2)

Or, [{-2+√(4-12k^2)}÷2k]+[{-2-√(4-12k^2)}÷2k ]=[{-2+√(4-12k^2)}÷2k}{-2-√(4-12k^2)}÷2k]

Or, {-2+√(4-12k^2) -2-√(4-12k^2)}÷2k= -1/4k^2{√(4-12k^2)-2}{√(4-12k^2)+2}

Or, {-2-2}÷2k= -1/4k^2[{√(4-12k^2)}^2 - 2^2}]

Or, -4÷2k= -1/4k^2(4-12k^2 - 4)

Or, -2/k= (-1/4k^2)(-12k^2)

Or, -2/k=-1/4×-12

Or, -2/k= 3

Therefore, k= -2/3