Ram and Shyam attempted to solve a quadratic equation. Ram made a mistake in writing down the constant term. He ended up with the roots (4, 3). Shyam made a mistake in writing down the coefficient ofx. He got the root as (3, 2). What will be the exact roots of the original quadratic equation?a)(6, 1)b)(-3, -4)c)(4, 3)d)(-4, -3)e)(-4, 3)Correct answer is option 'A'. Can you explain this answer?

Question

Answer

To find the exact roots of the original quadratic equation, we need to determine the correct constant term and coefficient of x.

Let's assume the quadratic equation is of the form ax^2 + bx + c = 0.

Mistake by Ram:
Ram made a mistake in writing down the constant term. He got the roots as (4, 3).
The sum of the roots of a quadratic equation is given by the formula:
Sum of roots = -b/a
So, 4 + 3 = -b/a
7 = -b/a

Mistake by Shyam:
Shyam made a mistake in writing down the coefficient of x. He got the root as (3, 2).
The product of the roots of a quadratic equation is given by the formula:
Product of roots = c/a
So, 3 * 2 = c/a
6 = c/a

Now, let's solve these two equations simultaneously to find the correct values of b and c.

Equation 1: 7 = -b/a
Equation 2: 6 = c/a

From Equation 1, we can write -b = 7a. (Multiplying both sides by -1)
Substituting this value of -b in Equation 2, we get 6 = (7a)(a)
6 = 7a^2
a^2 = 6/7
a = ±√(6/7)

Now, we can substitute the value of a in Equation 1 to find the value of b.
-(-b) = 7(±√(6/7))
b = 7(±√(6/7))

Therefore, the original quadratic equation is:
(±√(6/7))x^2 + 7(±√(6/7))x + 6 = 0

Simplifying this equation gives us:
(√(6/7))x^2 + 7(√(6/7))x + 6 = 0 or
(-√(6/7))x^2 - 7(√(6/7))x + 6 = 0

These two equations have the same roots, but different signs. So, the exact roots of the original quadratic equation are:
x = -7(√(6/7)) and x = -√(6/7) or
x = 7(√(6/7)) and x = √(6/7)

These roots can be simplified as:
x = -7√(6/7) and x = -√(6/7) or
x = 7√(6/7) and x = √(6/7)

Among these options, the correct answer is option 'A': (6, 1).

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