If both p and q belong to the set {1, 2, 3, 4}, then how many equations of the form px2 + qx + 1 = 0 will have real roots?

If both p and q belong to the set {1, 2, 3, 4}, then how many equations of the form px2 + qx + 1 = 0 will have real roots? Correct Answer 7

Concept:

For any quadratic equation, ax² + bx + c = 0. We have discriminant, D = b² - 4ac then the given quadratic equation has:

I. Distinct and real roots if D > 0.

II. Real and repeated roots, if D = 0.

III. Complex roots and conjugate of each other, D < 0.

Calculation:

Given:

Both p and q belong to the set {1, 2, 3, 4} and the quadratic equation of the form, px² + qx + 1 = 0 has real roots.

By comparing the quadratic equation of the form, px² + qx + 1 = 0 with the standard quadratic equation ax² + bx + c = 0. We get, a = p, b = q and c = 1.

⇒ D = b² - 4ac = q² - 4p

∵ The roots are real.

⇒ D ≥ 0 ⇒ q² ≥ 4p

∵ p and q belong to the set {1, 2, 3, 4}

Case -1:

If p = 1 then q² ≥ 4 ⇒ q ∈ {2, 3, 4}

Hence, for p = 1, q can take 3 values

So, 3 quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 1.

Case -2:

If p = 2 then q² ≥ 8 ⇒ q ∈ {3, 4}

Hence, for p = 2, q can take 2 values

So, 2 quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 2.

Case -3:

If p = 3 then q² ≥ 12 ⇒ q ∈ {4}

Hence, for p =1, q can take 1 value

So, only 1 quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 3.

Case -4:

If p = 4 then q² ≥ 16 ⇒ q ∈ {4}

Hence, for p =1, q can take 1 value

So, only 1 quadratic equation of the form, px² + qx + 1 = 0 can be formed for p = 4.