Class 10 Maths Revision Notes for Quadratic Equations of Chapter 4 - Free PDF Download
Quadratic Equations Class 10 Notes CBSE Maths Chapter 4 (Free PDF Download)
FAQs on Quadratic Equations Class 10 Notes CBSE Maths Chapter 4 (Free PDF Download)
1. What is the value of k for a quadratic equation (x - a) (x - 10) + 1 = 0. This equation has integral roots.
We can write the given quadratic equation as x2 - (10 + k) x + 1 + 10k = 0
This means that, D = b2 - 4ac = 100 + k2 + 20k - 40k = k2 - 20k + 96 = (k - 10)2 - 4
This quadratic equation will have integral roots if the value of the discriminant > 0. This means that D is a perfect square and a = 1, while b and c are integers
Hence, (k - 10)2 - D = 4
We know that the discriminant is a perfect square. This further means that the difference between the two perfect squares will be 4 only if D = 0 and (k - 10)2 = 4
Therefore, k - 10 = ± 2. And the values of k = 8 and 12
2. What is the value of k when the equation p / (x + r) + q / (x - r) = k / 2x has two equal roots?
We can write the given quadratic equation as:
[2p + 2q - k] x2 - 2r [p - q] x + r2k = 0
In case of equal roots, the discriminant or D = 0
This means that b2 - 4ac = 0
In this equation, a = [2p + 2q - k], b = -2r [p - q], and c = r2k
= [-2r (p - q)]2 - 4 (2p + 2q - k) (r2k)] = 0
R2 (p - q)2 - r2k (2p + 2q - k) = 0
We also know that r is not equal to zero.
Hence, (p - q)2 - k (2p + 2q - k) = 0
K2 - 2(p + q) k + (p - q)2
K = 2 (p + q) ± √[4 (p + q)2 - 4 (p - q)]2 / 2 = -(p + q) ± √4pq
Therefore, we can conclude that the value of k is (p + q) ± 2 √p x q = (√p ± √q)2