Question

If a, b, c are rational, prove that the roots of the equation (6 - c)x2 + (c - a)x + (a - b) = 0 are also rational.​

Answer 1

Step-by-step explanation:

On comparing the given equation with Ax

2

+Bx+C=0 we get,

A=(b+c−a) ; B=(c+a−b) and C=(a+b−c)

We know, D=B

2

−4AC

∴D=(c+a−b)

2

−4(b+c−a)(a+b−c)

=(a+b+c−2b)

2

−4(a+b+c−2a)(a+b+c−2c)

=(−2b)

2

−4(−2a)(−2c)=4(b

2

−4ac)

=4[(−a−c)

2

−4ac]=4(a−c)

2

={2(a−c)}

2

= Perfect square

∴ The roots are rational.