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If the quadratic equations x2 + ax + b = 0 and x2 + bx + a = 0 (a ≠ b) have a common root, then the numerical value of a + b is . . . . . . . . . . . . . . .

 If the quadratic equations x2 + ax + b = 0 and x2 + bx + a = 0 (a ≠ b) have a common root, then the numerical value of a + b is . . . . . . . . . . . . . . .

Grade:11

2 Answers

Kevin Nash
askIITians Faculty 332 Points
7 years ago
According to the question, x2 + ax + b = 0 and x2 + bx + a = 0 (a ≠ b) have a common root, let it be α then we have
α 2 + 2 α + b = 0 . . . . . . . . . . . . (i)
α2 + b α + a = 0 . . . . . . . . . . . . (ii)
⇒ α2/a2 -b2 = α/b – a = 1/b – a ⇒ α2/a + b = α/-1 = 1/-1
⇒ α2 = - (a + b); α = 1 ⇒ a + b = -1
∴ Numerical value of a + b = -1
abhay
11 Points
4 years ago
let the common root be z.
So,  z2+az+b=0  and z2+bz+a=0  as per the given equations.
Equate them , z2+az+b=z2+bz+a
                        (a-b)z=a-b
                         so, z=1
put z in any of the 2 equations,
12 + b.1+a=0    or 12 +a.1+b=0
1+a+b=0  or  1+a+b=0
so from both the cases (a+b)=-1
 
 
 
 

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