The number of roots common between the two eq...
The number of roots common between the two equations x3 + 3x+ 4x + 5 = 0 and x3+ 2x+ 7x + 3 = 0 is
• a)
0
• b)
1
• c)
2
• d)
3
The number of roots common between the two equations x3 + 3x2+ 4x + 5 ...
Subtracting one equation from other we get a quadratic equation x- 3x + 2 = 0, which has two roots 1,2, but neither of them satisfies these equations.
Hence, 0 will be the answer.
The number of roots common between the two equations x3 + 3x2+ 4x + 5 ...
Solution:

To find the number of roots common between the two equations, we need to find their intersection points.

Let the roots of the first equation be a, b, and c. Then, we can write the equation as:

(x-a)(x-b)(x-c) = 0

Expanding this, we get:

x3 - (a+b+c)x2 + (ab+bc+ca)x - abc = 0

Comparing this with the first equation, we get:

a+b+c = 3

ab+bc+ca = 4

abc = -5

Similarly, for the second equation, let the roots be p, q, and r. Then, we can write the equation as:

(x-p)(x-q)(x-r) = 0

Expanding this, we get:

x3 - (p+q+r)x2 + (pq+qr+rp)x - pqr = 0

Comparing this with the second equation, we get:

p+q+r = 2

pq+qr+rp = 7

pqr = -3

To find the common roots, we need to solve these two sets of equations simultaneously.

Adding the two sets of equations, we get:

a+b+c+p+q+r = 5

ab+bc+ca+pq+qr+rp = 11

abc+pqr = -8

Substituting the values of a+b+c and p+q+r from the above equations, we get:

5 - (ab+bc+ca) + 2 - (pq+qr+rp) = 0

Simplifying this, we get:

ab+bc+ca+pq+qr+rp = 7

Substituting this in the equation abc+pqr = -8, we get:

abc = -5

pqr = -3

Since the product of the roots is different for the two equations, they do not have any common roots.

Hence, the correct answer is option A.
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The number of roots common between the two equations x3 + 3x2+ 4x + 5 = 0 and x3+ 2x2+ 7x + 3 = 0 isa)0 b)1 c)2 d)3Correct answer is option 'A'. Can you explain this answer?
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