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Hint: In this question first of all we have to know the basics of algebra like in this question we should know that roots of equations are equal to zero if put them in variables by this concept we will able to solve half the question then we should also know the elimination method of linear equations in two variables also we should know about the relation of zeros in cubic polynomial by these simple things you should be able to find the solution.

Complete step-by-step answer:

To find ‘m’ and ‘n’, we will put the values of zeros of the equation in the place of ‘x’ and then equate the equation to zero.

$ \Rightarrow $ 2 and 3 are the zeros of the given equation $2{x^2} + mx - 13x + n$

Case 1

$

f(2) = 2{(2)^3} + m{(2)^2} - 13(2) + n = 0 \\

\Rightarrow 16 + 4m - 26 + n = 0 \\

\Rightarrow 4m + n = 10{\text{ }}......\left( 1 \right) \\

$

Case 2

$

f\left( 3 \right) = 2{(3)^3} + m{\left( 3 \right)^2} - 13\left( 3 \right) + n = 0 \\

\Rightarrow 54 + 9m - 39 + n = 0 \\

\Rightarrow 9m + n = - 15{\text{ }}.......{\text{(2)}} \\

$

Using (1) and (2) we get

$

4m + n = \;{\text{ }}10 \\

9m + n = - 15 \\

( - )( - ){\text{ ( + )}} \\

\\

$

$ - 5m{\text{ }} = {\text{ }}25{\text{ }}$

$

m = \dfrac{{25}}{{ - 5}} \\

m = - 5 \\

$

Put value of m in (1)

$

\Rightarrow 4m + n = 10 \\

\Rightarrow 4( - 5) + n = 10 \\

\Rightarrow - 20 + n = 10 \\

\Rightarrow n = 30 \\

$

Therefore, m=-5 and n=30

Now we know that $\alpha + \beta + \gamma = - \dfrac{b}{a}$ (sum of cubic roots)

$

2 + 3 + p = \dfrac{5}{2} \\

5 + p = \dfrac{5}{2} \\

p = \dfrac{5}{2} - \dfrac{5}{1} \\

p = \dfrac{{5 - 10}}{2} \\

p = \dfrac{{ - 5}}{2} \\

$

So, third root= $\dfrac{{ - 5}}{2}$

To determine m+n+2(p)

$

= - 5 + 30 + 2\left( { - \dfrac{5}{2}} \right) \\

= 25 - 5 \\

= 20 \\

$

Note: In this question we should know the basics of algebra like values of roots of equations if put instead of variables gives us 0. Also we must know about the elimination method of linear equations in two variables which will help us to find m and n. Then we must also be aware of the relation between the zeros of a cubic polynomial. Last but not the least we must know about the determination of the values and what are the values for cubic relations of equations. By these simple tips the solution becomes quite easier to calculate.

Complete step-by-step answer:

To find ‘m’ and ‘n’, we will put the values of zeros of the equation in the place of ‘x’ and then equate the equation to zero.

$ \Rightarrow $ 2 and 3 are the zeros of the given equation $2{x^2} + mx - 13x + n$

Case 1

$

f(2) = 2{(2)^3} + m{(2)^2} - 13(2) + n = 0 \\

\Rightarrow 16 + 4m - 26 + n = 0 \\

\Rightarrow 4m + n = 10{\text{ }}......\left( 1 \right) \\

$

Case 2

$

f\left( 3 \right) = 2{(3)^3} + m{\left( 3 \right)^2} - 13\left( 3 \right) + n = 0 \\

\Rightarrow 54 + 9m - 39 + n = 0 \\

\Rightarrow 9m + n = - 15{\text{ }}.......{\text{(2)}} \\

$

Using (1) and (2) we get

$

4m + n = \;{\text{ }}10 \\

9m + n = - 15 \\

( - )( - ){\text{ ( + )}} \\

\\

$

$ - 5m{\text{ }} = {\text{ }}25{\text{ }}$

$

m = \dfrac{{25}}{{ - 5}} \\

m = - 5 \\

$

Put value of m in (1)

$

\Rightarrow 4m + n = 10 \\

\Rightarrow 4( - 5) + n = 10 \\

\Rightarrow - 20 + n = 10 \\

\Rightarrow n = 30 \\

$

Therefore, m=-5 and n=30

Now we know that $\alpha + \beta + \gamma = - \dfrac{b}{a}$ (sum of cubic roots)

$

2 + 3 + p = \dfrac{5}{2} \\

5 + p = \dfrac{5}{2} \\

p = \dfrac{5}{2} - \dfrac{5}{1} \\

p = \dfrac{{5 - 10}}{2} \\

p = \dfrac{{ - 5}}{2} \\

$

So, third root= $\dfrac{{ - 5}}{2}$

To determine m+n+2(p)

$

= - 5 + 30 + 2\left( { - \dfrac{5}{2}} \right) \\

= 25 - 5 \\

= 20 \\

$

Note: In this question we should know the basics of algebra like values of roots of equations if put instead of variables gives us 0. Also we must know about the elimination method of linear equations in two variables which will help us to find m and n. Then we must also be aware of the relation between the zeros of a cubic polynomial. Last but not the least we must know about the determination of the values and what are the values for cubic relations of equations. By these simple tips the solution becomes quite easier to calculate.

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