If quadratic equation x2+2(a+2 b) x+(2 a+b-1)=0 has unequal real roots for all b ∈ R then the possible values of a can be equal to

Question

If quadratic equation x2+2(a+2 b) x+(2 a+b-1)=0 has unequal real roots for all b ∈ R then the possible values of a can be equal to (A) 5 (B) - 1 (C) - 10 (D) 3

Tardigrade Admin · Accepted Answer

The quadratic equation x 2+2( a +2 b )+(2 a + b -1)=0 will have unequal real roots, if D=4(a+2 b)2-4(2 a+b-1)>0 ⇒ a2+4 b2+4 a b>2 a+b-1 ⇒ 4 b2+(4 a-1) b+(a2-2 a+1)>0 ∀ b ∈ R ∴ (4 a-1)2-16(a2-2 a+1)<0 ⇒ (16 a2-8 a+1)-(16 a2-32 a+16)<0 ⇒ 24 a-15<0 Hence a< (15/24). Now verify alternatives.

Q. If quadratic equation x2+2(a+2b)x+(2a+b−1)=0 has unequal real roots for all b∈R then the possible values of a can be equal to

5

- 1

- 10

3

The quadratic equation x2+2(a+2b)+(2a+b−1)=0 will have unequal real roots, if D=4(a+2b)2−4(2a+b−1)>0 ⇒a2+4b2+4ab>2a+b−1 ⇒4b2+(4a−1)b+(a2−2a+1)>0∀b∈R ∴(4a−1)2−16(a2−2a+1)<0 ⇒(16a2−8a+1)−(16a2−32a+16)<0 ⇒24a−15<0 Hence a<2415​. Now verify alternatives.

Q. If quadratic equation $x^{2} + 2 (a + 2 b) x + (2 a + b - 1) = 0$ has unequal real roots for all $b \in R$ then the possible values of a can be equal to