It is given that the difference between the zeroes of 4x2 – 8kx + 9

1.	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» (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)

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»

(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)