Question

If \[2{{x}^{2}}+mxy+3{{y}^{2}}-5y-2\] have two rational factors then m is equal to
\[\left( a \right)\pm 7\]
\[\left( b \right)\pm 6\]
\[\left( c \right)\pm 5\]
\[\left( d \right)None\]

Answer 1

Hint:To solve this question, we will first calculate the determinant D of the given equation by considering the equation quadratic in terms of y. The discriminant of an equation is of the form \[a{{x}^{2}}+bx+c\] is given by \[D={{b}^{2}}-4ac.\] Then we want D to be a perfect square and that can be obtained by using \[{{b}^{2}}=4ac\] to get the result.

Complete step by step answer:
We are given \[2{{x}^{2}}+mxy+3{{y}^{2}}-5y-2......\left( i \right)\]
Consider x and y terms, we have,
\[3{{y}^{2}}+mxy-5y+2{{x}^{2}}-2\]
\[\Rightarrow 3{{y}^{2}}+y\left( mx-5 \right)+2{{x}^{2}}-2\]
Consider only the y as the variable and x as some number. Then the above obtained equation is quadratic (degree 2) equation in y. Now, we are given that \[2{{x}^{2}}+mxy+3{{y}^{2}}-5y-2\] has two rational factors.
The discriminant D of an equation \[a{{x}^{2}}+bx+c\] is given by \[D={{b}^{2}}-4ac.\] We have the equation as
\[\Rightarrow 3{{y}^{2}}+y\left( mx-5 \right)+2{{x}^{2}}-2\]
Then as the above equation is considered quadratic in y, therefore we get discriminant D as
\[D={{\left( mx-5 \right)}^{2}}-4\times \left( 3 \right)\left( 2{{x}^{2}}-2 \right)\]
\[\Rightarrow D={{m}^{2}}{{x}^{2}}+25-10mx-12\left( 2{{x}^{2}}-2 \right)\]
\[\Rightarrow D={{m}^{2}}{{x}^{2}}+25-10mx-24{{x}^{2}}+24\]
Now, clubbing similar terms together, we have
\[\Rightarrow D={{x}^{2}}\left( {{m}^{2}}-24 \right)-10mx+49\]
Now 10mx can be written as
\[10mx=2\times 7\times \dfrac{5}{7}mx\]
Using this in the above equation, we have,
\[\Rightarrow D={{x}^{2}}\left( {{m}^{2}}-24 \right)-2\times 7\times \dfrac{5}{7}mx+{{7}^{2}}\]
We need D to be a perfect square. So, D will be a perfect square if \[{{x}^{2}}\left( {{m}^{2}}-24 \right)-2\times 7\times \dfrac{5}{7}mx+{{7}^{2}}\] is in the form of \[{{b}^{2}}=4ac.\] And here the equation changes in terms of x and not y. Then,
\[b=2\times 7\times \dfrac{5}{7}m\]
\[a=\left( {{m}^{2}}-24 \right)\]
\[c={{7}^{2}}\]
Then substituting the values of a, b and c, we have,
\[\Rightarrow {{b}^{2}}=4ac\]
\[\Rightarrow {{\left( 2m\times 7\times \dfrac{5}{7} \right)}^{2}}=4\times \left( {{m}^{2}}-24 \right)\times {{7}^{2}}\]
\[\Rightarrow {{\left( 10 \right)}^{2}}{{m}^{2}}=4\times 49\left( {{m}^{2}}-24 \right)\]
\[\Rightarrow \dfrac{100}{4}{{m}^{2}}=49\left( {{m}^{2}}-24 \right)\]
\[\Rightarrow {{m}^{2}}-24=\dfrac{100}{4\times 49}{{m}^{2}}\]
\[\Rightarrow {{m}^{2}}-24=\dfrac{25}{49}{{m}^{2}}\]
Taking the terms in m to one side, we have,
\[\Rightarrow {{m}^{2}}-\dfrac{25}{49}{{m}^{2}}=24\]
\[\Rightarrow {{m}^{2}}\left( 1-\dfrac{25}{49} \right)=24\]
\[\Rightarrow {{m}^{2}}\left( \dfrac{49-25}{49} \right)=24\]
\[\Rightarrow {{m}^{2}}\left( \dfrac{24}{49} \right)=24\]
\[\Rightarrow {{m}^{2}}=49\]
Taking square root on both the sides, we get,
\[\Rightarrow m=\sqrt{49}\]
\[\Rightarrow m=\pm 7\]
Therefore, the two rational factors of m are \[\pm 7.\]
Hence, option (a) is the right answer.

Note:
A possible confusion can be at the point where \[{{b}^{2}}-4ac\] or \[{{b}^{2}}=4ac\] is considered as multiple or two times. Firstly, we have formed the equation is quadratic in terms of y, then we have used \[{{b}^{2}}-4ac=D\] where D is the discriminant. Then after obtaining D, we have again used \[{{b}^{2}}=4ac\] for a perfect square. Then we have used x as equation variable.