Knowing that 2 and 3 are the roots of the equation $2{x^3}$$ + {\text{ m}}{x^2}$ - $13x{\text{ + n = 0}}$, determine m + n +2(p)? Where p is the root of the given equation.
Last updated date: 26th Mar 2023
•
Total views: 306.9k
•
Views today: 2.83k
Answer
306.9k+ views
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.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India
