Question

2. Find the values of k for each of the following quadratic equations, so that they have two
equal roots.
(1) 2x^2 + kx+ 3 = 0
=
(ii) kx (x - 2) +6=0
​

Answer 1

Answer:

(1)we have to find the values of k for quadratic equations 2x² + kx + 3 = 0 so that they have two equal roots.

we know, quadratic equation will be equal only when

discriminant, D = b² - 4ac = 0

on comparing 2x² + kx + 3 = 0 with general form of quadratic equation , ax² + bx + c = 0 we get, a = 2, b = k and c = 3

so Discriminant , D = (k)² - 4(2)(3) = 0

or, k² - 24 = 0

or, k = ± √24 = ±2√6

hence, the value of k = 2√6 or -2√6

(2) Kx^2-2kx +6=0

Equal roots implies that D=0

This implies that b^2-4ac=0

(-2k)^2 -24k=0

4k^2-24k=0

4k(k-6)=0

4k=0

This implies that k=0

And k-6=0

This implies that k=6

So,k=0,6

Answer 2

Answer:

k=4

Step-by-step explanation:

By cross - multiplication method

2×6=3k

12=3k

k=4