If the equations x 2 + b x - 1 = 0 and x 2 + x + b = 0 have a common root different from - 1 , then b is…

If the equations x2+bx-1=0 and x2+x+b=0 have a common root different from -1, then b is equal to :
  1. 2
  2. 3
  3. 3
  4. 2

Solution

Let common root be x
x2+bx-1=0     ...1

x2+x+b=0       ...2

1-2, we get x=b+1b-1

Put x in Equation 1, we get

b+1b-12+b+1b-1+b=0

⇒ b+12 + b+1 b-1+bb-12=0

⇒ b2+1+2b+b2-1+bb2-2b+1=0

⇒ 2b2+2b+b3-2b2+b=0

⇒  b3+3b=0
⇒   bb2+3=0⇒b=0 or b2=-3

When  b=0 then common root x= -1
But the given common root x≠-1

∴ b2= -3

b= ± 3i

b= 3 Alternative Solution: Theek hai, chalo step by step isko samajhte hain. Dekho, humare paas do equations hain: 1) $x^2 + bx - 1 = 0$ 2) $x^2 + x + b = 0$ Ab humein pata hai ki dono equations ka ek common root hai jo -1 se alag hai. Let's say ki yeh common root hai $p$. Toh ab hum $p$ ko dono equations mein substitute karke dekhte hain. First equation mein daalte hain: $p^2 + bp - 1 = 0$ Ab second equation mein daalte hain: $p^2 + p + b = 0$ Ab hum in dono equations ko subtract karenge. Dekho, $p^2$ toh cancel ho jayega kyuki dono equations mein same hai. Toh humare paas bachega: $bp - p - (b+1) = 0$ Isko thoda aur simplify karke dekhte hain: $p(b - 1) - (b + 1) = 0$ Ab hum $p$ ko isolate karte hain: $p = \frac{b+1}{b-1}$ Ab dhyan se dekho, humare paas do roots honge dono equations ke. Ek toh $p$ hai, aur dusra hai $-1$. Agar $-1$ root nahi hai, toh $p$ ka value $-1$ se alag hoga. Toh hum $p$ ki value $-1$ se compare karenge: $\frac{b+1}{b-1} ≠ -1$ Isko solve karne par, humein $b = \pm \sqrt{3}$ milta hai. Lekin humare options mein sirf $+\sqrt{3}$ hai, toh wahi hoga humara answer. Toh samjha? Bas itna hi! Keep practicing, and you'll get better at this.

Asked in: JEE Main 2016 (09 Apr Online)

Practice more Quadratic Equation questions on Aicharya