An equation has been given:
$2x^2 +5xy + 3y^2 +7y -98=0$
It has been said that it represents two straight lines. What are they?
MY ATTEMPT:
I broke up $5xy$ into $2xy$ and $3xy$. So I got-
$(x+y)(2x+3y) + 7y - 98=0$
But this isn't right as pair of straight lines is the product of equations of two straight lines. So, how should I break this up.Also, what is the general method of finding the equations of two straight lines when their combined equation has been provided?
On
$$2x^2 +5xy + 3y^2 +7y -98=(x+y+7)(2x+3y-14)$$
One way to find this factorization is to write
$$(ax+by+c)(dx+ey+f)$$
and then equate coefficients. Here we can see that
$$ad= 2,\quad be=3,\quad ae+bd=5,\quad af+cd=0,\quad bf+ce=7\quad cf=-98$$
On
You need to factorize it. Given the existance of squared terms as well as a constant term it is likely to be of the form:
$$(ax+by+c)(dx+ey+f)$$
Then you can apply techniques/thinking similar to factorizing quadratics. Also as its been set as homework I'm going to assume integer coefficients.
$ad=2$ gives $a=1$ and $d=2$ without loss of generality.
The $xy$ term gives that $5=ae+bd=e+2d$. Combining this with $be=3$ and limiting values to integers gives $e=3$ and $b=1$.
Lastly $0=af+dc=f+2c$ so $f=-2c$. Combining with $cf=-98$ we get $c^2=49$. Checking signs leads to $c=7$ and $f=-14$.
So the factorized form is: $(x+y+7)(2x+3y-14)$.
On
See $$2x^2+(5y)x+(3y^2+7y-98)=0$$ as a quadratic equation in $x$.
First, note that $$3y^2+7y-98=(3y-14)(y+7)$$ Second, we can have $$5y=1\cdot (3y-14)+2\cdot (y+7)$$ Using these,
$$\begin{align}2x^2+5xy+3y^2+7y-98&=2x^2+5xy+(3y-14)(y+7)\\&=2x^2+(3y-14+2(y+7))x+(3y-14)(y+7)\\&=2x^2+(3y-14)x+2(y+7)x+(3y-14)(y+7)\\&=2x(x+y+7)+(3y-14)(x+y+7)\\&=(x+y+7)(2x+3y-14)\end{align}$$
In general, $$abx^2+(ad+bc)x+cd=(ax+c)(bx+d)$$ In our case, $$a=1,\quad b=2,\quad c=y+7,\quad d=3y-14.$$
On
This might be getting ahead of where you are in your studies, but here’s another way to proceed once you’ve factored the quadratic part of the equation.
After factoring the equation of the two lines, you know that their individual equations will be of the form $x+y=c_1$ and $2x+3y=c_2$, respectively. So, you just have to find those two constants and you’re done. If a line is given by the equation $ax+by=c$ and you don’t know what $c$ is, you can find it by plugging in a known point. One possibility is the intersection point of the two lines.
If you treat the left-hand side of the equation as a function of two variables, it turns out that the two lines intersect at the point at which the function’s gradient is zero. We have $$\nabla(2x^2 +5xy + 3y^2 +7y -98)=\langle4x+5y,5x+6y+7\rangle,$$ so the intersection point is the solution of the system of linear equations $$\begin{align}4x+5y&=0\\5x+6y&=-7\end{align}$$ which you can find to be $x=-35$, $y=28$. Plugging this point into the individual line equations above gives $c_1=-7$ and $c_2=14$. You can verify this by expanding $(x+y+7)(2x+3y-14)=0$. You should get your original equation back.
Starting with the factorization $$ (2x+3y)(x+y)=2x^2+5xy+3y^2 $$ we can proceed with $$ \begin{align} (2x+3y+a)(x+y+b) &=2x^2+5xy+3y^2+a(x+y)+b(2x+3y)+ab\\ &=2x^2+5xy+3y^2+7y-98 \end{align} $$ where $a=-2b$ to eliminate the $x$ term and $b=7$ to get the $7y$ term.
Therefore, we get $$ (2x+3y-14)(x+y+7)=2x^2+5xy+3y^2+7y-98 $$