If $x^{2} - 6 x + 5 = 0$ and $x^{2} - 12 x + p = 0$ have a common root, then find the value of p.

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11 and 35

11 and 385

-11 and -35

None of the Above

answer is A.

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Detailed Solution

Given that

x^{2} - 6 x + 5 = 0

and

x^{2} - 12 x + p = 0

have common roots.
And we know that if two quadratic equations have one common root then the condition which is

{(a}_{1}^{} c_{2}^{} - {a_{2}^{} c_{1})}^{2} = {(a}_{1}^{} b_{2}^{} - {a_{2}^{} b_{1})}^{2} {(b}_{1}^{} c_{2}^{} - {b_{2}^{} c_{1})}^{2}

is satisfied by the coefficients of the given equations.
So, now we will put the coefficients of the given two equations in the above-given relation and solve the equation to get the value of p.
Now, by comparing with the general form of a quadratic equation we get the values of the coefficient as:

a_{1} = 1, b_{1} = - 6, c_{1} = 5, a_{2} = 1, b_{2} = - 12, c_{2} = p

Now, by putting the value of the coefficient of the given two equations we get,

\Rightarrow (1 \times p - 15)^{2} = {{1 \times (- 12) - 1 \times (- 6)}^{2}} {{(- 6) \times p - (- 12) \times 5}^{2}}

\Rightarrow (p - 5)^{2} = (- 12 + 6) (- 6 p + 60)

By, using the formula

(x - y)^{2} = (x^{2} + y^{2} + 2 xy)

{\Rightarrow p}^{2} + 25 - 10 p = (- 6) (- 6 p = 60)

{\Rightarrow p}^{2} + 25 - 10 p = 36 p - 360

{\Rightarrow p}^{2} - 46 p + 385 = 0

Now, we have to factorise the given equation. So break 46p into two parts such that their product is

{385 p}^{2}

{\Rightarrow p}^{2} - 11 p - 35 p + 385 = 0

\Rightarrow p (p - 11) - 35 (p - 11) = 0

\Rightarrow

(p - 11) (p - 35) = 0

This means, that

(p - 11) and (p - 35)

both are equal to zero. So p = 11 and p = 35.
Thus there are two possible values of p that is, 11 and 35 for which the given two equations have one common root.

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