Question

It is given that the difference between the zeroes of 4x² – 8kx + 9 is 4 and k > 0. Then, k = ?

(a) \(\frac{1}2\) 

(b) \(\frac{3}2\)

(c) \(\frac{5}2\)

(d) \(\frac{7}2\)

Answer 1

(c) \(\frac{5}2\)

Let the zeroes of the polynomial be α and α + 4 

Here, p(x) = 4x² – 8kx + 9 

Comparing the given polynomial with ax² + bx + c, 

we get: 

a = 4, b = -8k and c = 9

Now, sum of the roots = \(\frac{-b}a\)

⇒ α + α + 4 = \(\frac{-(-8)}4\)

⇒ 2α + 4 = 2k 

⇒ α + 2 = k 

⇒ α = (k – 2) ….(i) 

Also, product of the roots, αβ = \(\frac{c}a\)

⇒ α (α + 4) = \(\frac{9}4\)

⇒ (k – 2) (k – 2 + 4) = \(\frac{9}4\)

⇒ (k – 2) (k + 2) = \(\frac{9}4\)

⇒ k² – 4 = \(\frac{9}4\)

⇒ 4k² – 16 = 9 

⇒ 4k² = 25

⇒ k² = \(\frac{25}4\)

⇒ k = \(\frac{5}4\)

(∵ k >0)