Mr McLogan, I know that you're not a fan of adding fractions and multiplying both sides of a rational equation by the LCD, your go to technique, is a one size fits all sledgehammer that always works but sometimes being able to express each side as a single fraction is much more computationally efficient. Take this problem for example 3x/(x + 5) + 1/(x - 2) = 7/(x^2 + 3x -10). If you add the expressions on the left to get a single fraction, to get [3x(x-2) + (x + 5)]/[(x+5)(x-2)] = 7/[x^2 + 3x - 10]. FOILING the denominator on the left you get [3x(x-2) + (x + 5)]/[x^2 + 3x - 10] = 7/[x^2 + 3x - 10]. Since the denominators on each side are identical you may eliminate them by inspection. This leaves you with the equation 3x(x - 2) + x + 5 = 7. Subtracting 7 from both sides gives 3x(x - 2) + x - 2 = 0. You can factor this equation by grouping to get (3x + 1)(x - 2) = 0. x = 2 is not a solution because x - 2 is a factor of the LCD. Therefore x = - 1/3. Looks like a lot of work but it's not. I was able to solve it this way in seconds though it took me minutes to explain it. 🤔
There is also a method that allows you to calculate it using the group or box method. It can take longer than doing it in your head, but is more direct and possible to determine the factors through an algorithm.
@@meeraali9207 take the form ax^2 + bx +c 1) multiply ax2 by c 2) look for 2 numbers that multiply to that result and add to bc 3) split BX into those 2 numbers so now it's ax2 + b1x +b2x +c. 4)group ax2 and b1x. Find a common factor and factor it out. Group b2x and c. You should have two groups that are identical with leading coefficients in front of them. 5) factor out the binomial common factor. That will be the first factor group. And the lead coefficients become the second factor group. This will factor any trinomial that is factorable. I know that is really abstract. Hard to do an example on here lol
Mr McLogan, I know that you're not a fan of adding fractions and multiplying both sides of a rational equation by the LCD, your go to technique, is a one size fits all sledgehammer that always works but sometimes being able to express each side as a single fraction is much more computationally efficient.
Take this problem for example
3x/(x + 5) + 1/(x - 2) = 7/(x^2 + 3x -10).
If you add the expressions on the left to get a single fraction, to get
[3x(x-2) + (x + 5)]/[(x+5)(x-2)] = 7/[x^2 + 3x - 10]. FOILING the denominator on the left you get
[3x(x-2) + (x + 5)]/[x^2 + 3x - 10] = 7/[x^2 + 3x - 10].
Since the denominators on each side are identical you may eliminate them by inspection. This leaves you with the equation 3x(x - 2) + x + 5 = 7. Subtracting 7 from both sides gives 3x(x - 2) + x - 2 = 0. You can factor this equation by grouping to get (3x + 1)(x - 2) = 0. x = 2 is not a solution because x - 2 is a factor of the LCD. Therefore x = - 1/3.
Looks like a lot of work but it's not. I was able to solve it this way in seconds though it took me minutes to explain it. 🤔
Great
There is also a method that allows you to calculate it using the group or box method. It can take longer than doing it in your head, but is more direct and possible to determine the factors through an algorithm.
what is the method
@@meeraali9207 to be clear this method only works if it is a) factorable and b) is no more than 4 terms.
@@meeraali9207 take the form ax^2 + bx +c
1) multiply ax2 by c
2) look for 2 numbers that multiply to that result and add to bc
3) split BX into those 2 numbers so now it's ax2 + b1x +b2x +c.
4)group ax2 and b1x. Find a common factor and factor it out. Group b2x and c. You should have two groups that are identical with leading coefficients in front of them.
5) factor out the binomial common factor. That will be the first factor group. And the lead coefficients become the second factor group. This will factor any trinomial that is factorable. I know that is really abstract. Hard to do an example on here lol
First!