Given that for any rational number a, the quadratic equation 2x^2 +(a+1)x-(3a^2 -4a+b)=0 always has two rational solutions for x. Find the value of b.

Question

Working out please

Answer 1

We can use the quadratic formula to find the roots (zeroes), using just the expression under the square root sign. For rational zeroes we need a perfect square. B²-4AC must be positive for real zeroes and it must evaluate to a perfect square, where A=2, B=a+1, C=-(3a²-4a+b).

B²-4AC=a²+2a+1+8(3a²-4a+b)=

25a²-30a+8b+1.

This has to be a perfect square which is of the form (5a-d)²=25a²-10ad+d², where d has to be determined.

So, 10ad must be equal to 30a, making d=3. d²=9 so 8b+1=9, b=1. So 5a-3 is the square root which is rational when a is any rational number and b=1.