Question

If quadratic equation x2 + ax + b/4 = 0 has rational roots such that a and b are positive integers less than 6, then number of possible pairs of (a, b) is
a) 4
b)5
c)6
d)7

Answer 1

Answer:

a) 4

Step-by-step explanation:

Let us calculate the discriminant of the quadratic equation.

$D = {b}^{2} - 4ac \\ D = (2(a + 1)) ^{2} - 4a(a + 2) \\D = 4 {a}^{2} + 8a + 4 - 4 {a}^{2} - 8a \\ D = 4$

So, the roots of the given equation will be

$x = \frac{ - b ± \sqrt{d} }{2a } \\ x = \frac{ - 2(a + 1 ) ±2}{2(a + 2)} \\ x = - \frac{a}{a + 2} \: - 1$

As x = -1 is already an integral solution.

So, we will consider the other case.

$x = - \frac{a}{a + 2} \\ = \frac{a + 2 - 2}{a + 2} \\ = - 1 + \frac{2}{a + 2}$

For 2/a+2 to be an integer, the positive values of a are 0, -1, -3 and -4.

Hence, there are 4 integral values of a .

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