1.

Find value of k for which following quadratic equations have real and fractional roots.(i) kx2 + 2x + 1 = 0(ii) kx2 + 6x + 1 = 0(iii) x2 – kx + 9 = 0

Answer»

(i) Given roots kx2 + 2x + 1 = 0

Comparing it by general quadratic equation

ax2 + bx + c = 0

a = k, b = 2, c = 1

Discriminant (D) = b2 – 4ac

= (2)2 – 4 × k × 1

= 4 – 4 k

= 4(1 – k)

For real and fractional roots

Discriminant (D) > 0

⇒ 4(1 -k) > 0

⇒ 1 – k > 0

⇒ k < 1

Thus, k will be less than.

(ii) Given equation kx2 + 6x + 1 = 0

Comparing it by general quadratic equation ax2 + bx + c – 0,

a = k, b = 6, c = 1

Discriminant (D) = b2 – 4ac

= (6)2 – 4 × k × 1

= 36 – 4k

= 4(9 – k)

If discriminant (D) > 0 then roots of equation will be real and fractional.

i.e., D > 0

⇒ 4(9 – k) > 0

⇒ 9 – k > 0

⇒ k > 9

Thus value of k will be greater than 9

(iii) Given equation x2 – kx + 9 = 0

Comparing it by general quadratic equation

ax2 + bx + c = 0

a = 1, b = -k, c = 9

Discriminant (D) = b2 – 4ac

= (-k)2 – 4 × 1 × 9

= k2 – 36

If D > 0, then roots of equation will be real and fractional.

D > 0

⇒ k2 – 36 > 0

⇒ (k – 6)(k + 6) > 0

⇒ k – 6 > 0 or k + 6 > 0

⇒ k > 6 or k < -6

Thus, value of k will be greater than 6 or less than -6



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