# Find the values of ‘m’ for which \[{{x}^{2}}+3xy+x+my-m\] has two linear factors in x and y, with integer coefficients.

(a) m = 1 or m = 11

(b) m = 0 or m = 12

(c) m = 10 or m = 0

(d) None

Last updated date: 20th Mar 2023

•

Total views: 305.4k

•

Views today: 7.84k

Answer

Verified

305.4k+ views

Hint: First of all, the given equation with a general 2-degree equation in x and y that is \[a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0\]. Get the values of a, b, c, f, g and h and substitute these in condition to resolve the equation into linear factors that is in \[abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0\] to get the value of m.

Complete step-by-step answer:

We are given an expression \[{{x}^{2}}+3xy+x+my-m\] that has 2 linear factors in x and y with the integer coefficient. Here, we have to find the value of ‘m’.

We know that general two-degree equation in x and y is:

\[f\left( x,y \right)=a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0....\left( i \right)\]

We know that to resolve the above equation into linear factors of x and y, it must satisfy the following condition:

\[abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0....\left( ii \right)\]

Now, let us consider the expression given in the question, that is,

\[g\left( x,y \right)={{x}^{2}}+3xy+x+my-m\]

Let us equate it to 0 to find its linear factors. So, we get,

\[{{x}^{2}}+3xy+x+my-m=0\]

As we know that the above equation is also a 2-degree equation in x and y. So now, we compare the coefficients of the above equation with that of equation (i), we get

Value of a = 1

Value of b = 0

Value of 2h = 3 or \[h=\dfrac{3}{2}\]

Value of 2g = 1 or \[g=\dfrac{1}{2}\]

Value of 2f = m or \[f=\dfrac{m}{2}\]

Value of c = –m

By substituting these values of a, b, c, f, g and h in equation (ii), we get,

\[\left( 1 \right)\left( 0 \right)\left( -m \right)+2.\left( \dfrac{m}{2} \right)\left( \dfrac{1}{2} \right)\left( \dfrac{3}{2} \right)-1{{\left( \dfrac{m}{2} \right)}^{2}}-0{{\left( \dfrac{1}{2} \right)}^{2}}-\left( -m \right){{\left( \dfrac{3}{2} \right)}^{2}}=0\]

By simplifying the above equation, we get,

\[0+\dfrac{3m}{4}-\dfrac{{{m}^{2}}}{4}-0+\dfrac{9}{4}m=0\]

Or, \[\dfrac{3m}{4}-\dfrac{{{m}^{2}}}{4}+\dfrac{9}{4}m=0\]

By multiplying (– 4) on both sides of the above equation, we get,

\[{{m}^{2}}-9m-3m=0\]

Or, \[{{m}^{2}}-12m=0\]

By taking m common in the above equation, we get,

\[m\left( m-12 \right)=0\]

So, we get, m = 0 and m = 12.

Hence we get 2 values of m that are 0 and 12 such that the given expression would be resolved into linear factors of x and y.

Therefore, option (b) is correct.

Note: Here, we can cross-check if these values of m are resolving the given expression into the linear factors or not as follows:

Let us take our given expression, that is,

\[g\left( x,y \right)={{x}^{2}}+3xy+x+my-m=0\]

Let us substitute m = 0 in the above equation, we get,

\[{{x}^{2}}+3xy+x=0\]

By taking x common, we get,

\[\left( x \right)\left( x+3y+1 \right)=0\]

Hence, we get 2 linear factors as x and (x + 3y + 1). Similarly, we can check for m = 12.

Complete step-by-step answer:

We are given an expression \[{{x}^{2}}+3xy+x+my-m\] that has 2 linear factors in x and y with the integer coefficient. Here, we have to find the value of ‘m’.

We know that general two-degree equation in x and y is:

\[f\left( x,y \right)=a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0....\left( i \right)\]

We know that to resolve the above equation into linear factors of x and y, it must satisfy the following condition:

\[abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0....\left( ii \right)\]

Now, let us consider the expression given in the question, that is,

\[g\left( x,y \right)={{x}^{2}}+3xy+x+my-m\]

Let us equate it to 0 to find its linear factors. So, we get,

\[{{x}^{2}}+3xy+x+my-m=0\]

As we know that the above equation is also a 2-degree equation in x and y. So now, we compare the coefficients of the above equation with that of equation (i), we get

Value of a = 1

Value of b = 0

Value of 2h = 3 or \[h=\dfrac{3}{2}\]

Value of 2g = 1 or \[g=\dfrac{1}{2}\]

Value of 2f = m or \[f=\dfrac{m}{2}\]

Value of c = –m

By substituting these values of a, b, c, f, g and h in equation (ii), we get,

\[\left( 1 \right)\left( 0 \right)\left( -m \right)+2.\left( \dfrac{m}{2} \right)\left( \dfrac{1}{2} \right)\left( \dfrac{3}{2} \right)-1{{\left( \dfrac{m}{2} \right)}^{2}}-0{{\left( \dfrac{1}{2} \right)}^{2}}-\left( -m \right){{\left( \dfrac{3}{2} \right)}^{2}}=0\]

By simplifying the above equation, we get,

\[0+\dfrac{3m}{4}-\dfrac{{{m}^{2}}}{4}-0+\dfrac{9}{4}m=0\]

Or, \[\dfrac{3m}{4}-\dfrac{{{m}^{2}}}{4}+\dfrac{9}{4}m=0\]

By multiplying (– 4) on both sides of the above equation, we get,

\[{{m}^{2}}-9m-3m=0\]

Or, \[{{m}^{2}}-12m=0\]

By taking m common in the above equation, we get,

\[m\left( m-12 \right)=0\]

So, we get, m = 0 and m = 12.

Hence we get 2 values of m that are 0 and 12 such that the given expression would be resolved into linear factors of x and y.

Therefore, option (b) is correct.

Note: Here, we can cross-check if these values of m are resolving the given expression into the linear factors or not as follows:

Let us take our given expression, that is,

\[g\left( x,y \right)={{x}^{2}}+3xy+x+my-m=0\]

Let us substitute m = 0 in the above equation, we get,

\[{{x}^{2}}+3xy+x=0\]

By taking x common, we get,

\[\left( x \right)\left( x+3y+1 \right)=0\]

Hence, we get 2 linear factors as x and (x + 3y + 1). Similarly, we can check for m = 12.

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