Question on common root s of quadratic equations

Question

If 3 different polynomials x^{​​​​​​2}​​​+ax+b , x^{​​​​​​2}​​​​​x2+x+ab , and ax^{​​​​​2}​​​​ax2+x+b have exactly one common root , where a,b are non zero numbers , then prove that the value of a+2b=0

Answer 1

We are given three different polynomials:

1. P₁(x) = x² + ax + b
2. P₂(x) = x² + x + ab
3. P₃(x) = ax² + ax + b

These polynomials are said to have exactly one common root, where a and b are non-zero numbers. We are asked to prove that a + 2b = 0.

Step 1: Assume the common root is r

Let the common root be r. Therefore, we know that:

P₁(r) = 0 ⇒ r² + ar + b = 0 

P₂(r) = 0 ⇒ r² + r + ab = 0

P₃(r) = 0 ⇒ ar² + ar + b = 0

We now have the following system of equations:

1. r² + ar + b = 0
2. r² + r + ab = 0
3. ar² + ar + b = 0

Step 2: Subtract the equations to simplify

To eliminate r², we subtract the second equation from the first equation:

(r² + ar + b) − (r² + r + ab) = 0

Simplifying:

ar + b − r − ab = 0

(ar − r) + (b − ab) = 0

Factor out the common terms:

r(a−1) + b(1−a) = 0

r(a − 1) − ab + b = 0

r(a − 1) = ab − b

r(a − 1) = b(a − 1)

If a ≠ 1, we can divide both sides by a - 1 (which is non-zero):

r = b

Step 3: Substitute r = b into the equations

Now that we know r = b, we substitute r = b into one of the original equations. Let's use r² + ar + b = 0:

b² + ab + b = 0

Factor the equation:

b(b + a + 1) = 0

Since b ≠ 0, we must have:

b + a + 1 = 0

Thus,

a + b = -1

Step 4: Prove a + 2b = 0

From the equation a + b = −1, we can solve for a:

a = −1 − b

Now substitute this into the equation a + 2b = 0:

(-1 - b) + 2b = 0

Simplify:

−1 + b = 0

b = 1

Step 5: Find a

Substitute b = 1 into a = −1 − b:

a = −1 − 1 = −2

Step 6: Conclusion

We have found that a = −2  and b = 1. Therefore:

a + 2b = −2 + 2(1) = −2 + 2 = 0

Thus, we have proved that a + 2b = 0.