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# How do you factor quadratic equations with two variables?

Last updated date: 17th Jun 2024
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Hint:Factoring reduces the higher degree equation into its linear equation. In the above given question, we need to reduce the quadratic equation into its simplest form in such a way that addition of products of the factors of first term and last should be equal to the middle one.

Complete step by step solution:
$a{x^2} + bxy + c{y^2} + dx + ey + f = 0$ is a general way of writing quadratic equations where a, b c,d,e and f are the numbers where $a,c \ne 0$
Now we can take can example of an equation,
$2{x^2} + 7xy - 15{y^2}$
In the above expression,
a=2, b=7, c=-15 d=0 e=0 f=0
First step is by multiplying the term $2{x^2}$ and the constant term -15, we get $- 30{x^2}$.
After this, factors of $- 30{x^2}$ should be calculated in such a way that their addition should be equal to 7xy.
Factors of -30 can be -3 and 10
where $- 3xy + 10xy = 7xy$.
So, further we write the equation by equating it with zero and splitting the middle term according to the factors.
$\Rightarrow 2{x^2} + 7xy - 15{y^2} = 0 \\ \Rightarrow 2{x^2} - 3xy + 10xy - 15{y^2} = 0 \\ \\$
Now, by grouping the first and last two terms we get common factors.
$\Rightarrow \left( {2x - 3y} \right) + 5y\left( {2x - 3y} \right) = 0 \\ \\$
Taking x common from the first group and 1 common from the second we get the above equation.
We can further solve it we get,
$\Rightarrow \left( {x + 5y} \right)(2x - 3y) = 0 \\ \\$
So here we get the above reduced form.

Note: An important thing to note is that suppose there is a trinomial expression given which has leading coefficients having perfect squares then we can bring the terms by writing it in perfect square then using the formula of ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$.