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**Hint:**Consider a general quadratic equation. Using the factorization method, form two quadratic equations from two sets of roots obtained by Ramesh and Mahesh. Apply the conditions of wrong coefficients in both the cases and write the final equation with correct coefficients. Use factorization method to find roots of the equation.

* A polynomial of degree ‘n’ can have at most ‘n’ possible real roots.

* Factorization method: If ‘p’ and ‘q’ are the roots of a quadratic equation, then we can say the quadratic equation is \[(x - p)(x - q) = 0\]

i.e. \[{x^2} - (p + q)x + pq = 0\] on further solving.

**Complete step-by-step answer:**

We are given that both Ramesh and Mahesh commit mistakes in finding the roots and find the roots as 8, 2 and -9, -1 respectively.

Since, both found number of roots as 2

\[ \Rightarrow \]Degree of the polynomial is 2.

So, we have a polynomial where variable has highest power 2.

\[\therefore \]We consider a quadratic equation \[a{x^2} + bx + c = 0\] … (1)

Coefficient of \[{x^2}\]is ‘a’; coefficient of \[x\]is ‘b’; constant value is ‘c’.

We take two cases for each solution done by Ramesh and Mahesh respectively.

CASE 1: Ramesh

We are given that Ramesh commits mistake in constant term and find the roots of equation as 8 and 2.

Use factorization method to form a quadratic equation.

Roots of quadratic equation are 8 and 2.

The quadratic equation is\[(x - 8)(x - 2) = 0\]

Multiply the brackets to obtain a quadratic equation

\[ \Rightarrow x \times (x - 2) - 8 \times (x - 2) = 0\]

\[ \Rightarrow (x) \times (x) + (x) \times ( - 2) + ( - 8) \times (x) + ( - 8) \times ( - 2) = 0\]

\[ \Rightarrow {x^2} - 2x - 8x + 16 = 0\]

Add the terms with same coefficient

\[ \Rightarrow {x^2} - 10x + 16 = 0\]

Here value of \[a = 1,b = - 10,c = 16\]

Since, Ramesh commits mistake in calculating the constant term i.e. ‘c’

\[ \Rightarrow c \ne 16\]

Also, we get \[a = 1,b = - 10\] … (2)

CASE 2: Mahesh

We are given that Mahesh commits a mistake in the coefficient of x and find the roots of equation as -9 and -1.

Use factorization method to form a quadratic equation.

Roots of quadratic equation are -9 and -1.

The quadratic equation is\[(x - ( - 9))(x - ( - 1)) = 0\]

Since multiplication of two negative signs gives a positive sign.

\[ \Rightarrow (x + 9)(x + 1) = 0\]

Multiply the brackets to obtain a quadratic equation

\[ \Rightarrow x \times (x + 1) + 9(x + 1) = 0\]

\[ \Rightarrow (x) \times (x) + (x) \times (1) + (9) \times (x) + (9) \times (1) = 0\]

\[ \Rightarrow {x^2} + x + 9x + 9 = 0\]

Add the terms with same coefficient

\[ \Rightarrow {x^2} + 10x + 9 = 0\]

Here value of \[a = 1,b = 10,c = 9\]

Since, Mahesh commits mistake in calculating the coefficient of x i.e. ‘b’

\[ \Rightarrow b \ne 10\]

Also, we get \[a = 1,c = 9\] … (3)

From equations (2) and (3) we get

\[a = 1,b = - 10,c = 9\]

Substitute the values of a, b and c in equation (1)

\[ \Rightarrow {x^2} - 10x + 9 = 0\] … (4)

The initially correct quadratic equation is \[{x^2} - 10x + 9 = 0\]

Use factorization method to find roots of equation (4)

We can write equation (4) as

\[ \Rightarrow {x^2} - x - 9x + 9 = 0\]

Take x common from first two terms and -9 common from last two terms

\[ \Rightarrow x(x - 1) - 9(x - 1) = 0\]

Collect the factors

\[ \Rightarrow (x - 1)(x - 9) = 0\]

Equate each factor to zero to obtain the roots of the equation.

If \[(x - 1) = 0\]

Shift constant value to RHS

\[ \Rightarrow x = 1\]

\[\therefore x = 1\] is a root.

If \[(x - 9) = 0\]

Shift constant value to RHS

\[ \Rightarrow x = 9\]

\[\therefore x = 9\] is a root.

Thus, the correct roots of the equation are 9, 1

**So, the correct option is B.**

**Note:**Students might try to solve this by taking the roots as \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. Keep in mind this process will not yield our answer as it is difficult to rule out coefficients and constant in this formula.

Also, many students form wrong quadratic equations using the given roots as they don’t multiply step by step which causes mistakes in one of the signs in the equation.

Students might get confused with the coefficient of \[{x^2}\], we write the factors of the equation according to the roots given, if the factors are in fraction form then we get a coefficient of \[{x^2}\], otherwise the coefficient remains 1.

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