factorise quadratic equations

[davey2015]

hello guys

i have a little problem i havent ever done any quadratic equations before and finding it abit hard, cant seen to start the equation, Can i have any advise and guidance on the following 2 questions please?

(i): x^2 -10x +21 = 0

(ii): 3x^2 -11x -4 = 0

thank you.

 

[stapel]

i havent ever done any quadratic equations before and finding it abit hard...

Did they not cover this in class or in your textbook? If not, there are loads of lessons available online.

...i cant seen to start the equation, Can i have any advise and guidance on the following 2 questions please?

(i): x^2 -10x +21 = 0

(ii): 3x^2 -11x -4 = 0

Your subject line mentions factoring. Would it be correct to guess that the instructions for these two equations were something like "solve by factoring"? If so, then where are you getting stuck? Have you studied this sort of factoring at all? If not, try this listing of lessons. Then consider:

(i) What two numbers multiply to a positive 21 and add to a negative 10?

(ii) What two numbers multiply to a negative 12 and add to a negative 11?

If the instructions were something else, kindly please reply with that information. Either way, when you reply, please include a clear listing of your efforts so far. Thank you!

 

[davey2015]

is this correct:

x^2 -10x +21 = 0
(x + 3)(x + 7) = 0
x^2 + 7x + 3x + 21
x = 3, x = 7

only reason why i ask is i cannot get 10x with both positives. but its the only 2 numbers that work correctly in the equation.

 

[stapel]

is this correct:

x^2 -10x +21 = 0
(x + 3)(x + 7) = 0
x^2 + 7x + 3x + 21
x = 3, x = 7

Since +7x + 3x = +10x, not -10x, this cannot be correct. However, it's fairly obvious that you must be on the right track, so...

only reason why i ask is i cannot get 10x with both positives. but its the only 2 numbers that work correctly in the equation.

Really? You can get a positive product (the +21) ONLY if you multiply two positive numbers?