If one root is common between two quadratic e...
If one root is common between two quadratic equations, x2 + 4x + 6 and ax2 + bx+ c, then find the product of the roots of ax2 + bx+ c. (Given: a, b and c are natural numbers.)
• a)
4
• b)
3
• c)
-2
• d)
6
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If one root is common between two quadratic equations, x2 + 4x + 6 and...
Solution: Discriminant of (x2 + 4x + 6) < 0 The equation will not have real roots. i.e., x2 + 4x + 6 has complex roots and they are conjugates of each other. One of the roots is also a root of ax2 + bx + c. Conjugate roots always occur in pairs.
So, the product of the roots of (ax2 + bx + c) is same as that of (x2 + 4x + 6). v a = 1 , b = 4 and c = 6 The product = c/a = 6 Hence, option 4.
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If one root is common between two quadratic equations, x2 + 4x + 6 and...
To find the product of the roots of the quadratic equation ax^2 + bx + c, we need to first find the roots of the equation x^2 + 4x + 6. Let's solve this equation first.

Solving the equation x^2 + 4x + 6 = 0
Using the quadratic formula, the roots of this equation can be found using the formula:

x = (-b ± √(b^2 - 4ac))/(2a)

For the given equation, a = 1, b = 4, and c = 6.

x = (-4 ± √(4^2 - 4(1)(6)))/(2(1))
x = (-4 ± √(16 - 24))/2
x = (-4 ± √(-8))/2
x = (-4 ± 2√2i)/2
x = -2 ± √2i

So, the roots of the equation x^2 + 4x + 6 = 0 are -2 + √2i and -2 - √2i.

Now, let's find the product of the roots of the given quadratic equation ax^2 + bx + c.

Product of the roots = (root1) * (root2)
= (-2 + √2i) * (-2 - √2i)
= (-2)^2 - (√2i)^2
= 4 - 2i^2
= 4 - 2(-1)
= 4 + 2
= 6

Therefore, the product of the roots of the quadratic equation ax^2 + bx + c is 6.

Hence, the correct answer is option D) 6.
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If one root is common between two quadratic equations, x2 + 4x + 6 and ax2 + bx+ c, then find the product of the roots of ax2 + bx+ c. (Given: a, b and c are natural numbers.)a)4b)3c)-2d)6Correct answer is option 'D'. Can you explain this answer?
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