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a little unsure

guys sorry it's me again i was factorising
3x2 + 10x + 7 = 0

in this case, would it be appropriate to use the quadratic formula, or would it not work because 3x2 isn't exactly the same as ax
Original post by entertainmyfaith
guys sorry it's me again i was factorising
3x2 + 10x + 7 = 0

in this case, would it be appropriate to use the quadratic formula, or would it not work because 3x2 isn't exactly the same as ax


How can a quadratic formula "not work"?

Completing the square, factorising, and using the quadratic formula (which is a general result from completing the square) are all valid methods of solving a quadratic, so they all work.

You can use the quadratic formula if you want, but you can also easily factorise this quadratic too.
Reply 2
no you can use it you would just make a= three
Original post by RDKGames
How can a quadratic formula "not work"?

Completing the square, factorising, and using the quadratic formula (which is a general result from completing the square) are all valid methods of solving a quadratic, so they all work.

You can use the quadratic formula if you want, but you can also easily factorise this quadratic too.


i didn't mean the formula not working, more if it was appropriate for the method because i was just unsure if 3x2 + 10x +7 = 0 was equivalent to ax + bx + c = 0- does the squared notation affect it?
Original post by entertainmyfaith
i didn't mean the formula not working, more if it was appropriate for the method because i was just unsure if 3x2 + 10x +7 = 0 was equivalent to ax + bx + c = 0- does the squared notation affect it?


Your "ax+bx+c=0" is not a quadratic expression, so no they are no equivalent.

If you meant ax2+bx+c=0ax^2+bx+c=0 then yes of course they are equivalent since we just have a=3,b=10,c=7a=3, b=10, c=7
Your gonna use the quadratic formula to FACTORISE.......
Original post by Ex-pert
Your gonna use the quadratic formula to FACTORISE.......


Don't see how that's a crime. The formula would give her solutions x1,x2x_1,x_2 to the equation so all that is needed to be done is plug them in the form (xx1)(xx2)=0(x-x_1)(x-x_2)=0 which would give the factorised form.
(edited 6 years ago)
Original post by entertainmyfaith
guys sorry it's me again i was factorising
3x2 + 10x + 7 = 0

in this case, would it be appropriate to use the quadratic formula, or would it not work because 3x2 isn't exactly the same as ax


the answer is;

[solution removed]
(edited 6 years ago)
Original post by RDKGames
Don't see how that's a crime. The formula would give her solutions x1,x2x_1,x_2 to the equation so all that is needed to be done is plug them in the form (xx1)(xx2)=0(x-x_1)(x-x_2)=0 which would give the factorised form.


But there is another easier way
I dont see the reason why he is making it look so complicated
(edited 6 years ago)
Original post by Ex-pert
But there is another easier way
I dont see the reason why he is making it look so complicated


I agree, but if that's the way she wants to try first then she can go ahead and settle on it for the time being.
i tried factorising it but i couldn't do it so i just used the quadratic formula. it got me the right answer
Original post by entertainmyfaith
i tried factorising it but i couldn't do it so i just used the quadratic formula. it got me the right answer


ok
Original post by entertainmyfaith
i tried factorising it but i couldn't do it so i just used the quadratic formula. it got me the right answer


Thats what i want to see
Original post by entertainmyfaith
i tried factorising it but i couldn't do it so i just used the quadratic formula. it got me the right answer


To factorise something of the form ax2+bx+c=0ax^2+bx+c=0 you want to look for two numbers λ,μ\lambda, \mu with the following properties: λ+μ=b\lambda + \mu = b and λμ=ac\lambda \mu =ac

Then you take the form (x+λa)(x+μa)=0\displaystyle \left( x+\frac{\lambda}{a} \right) \left( x+\frac{ \mu}{a} \right)=0 and plug those in there for factorisation.

In your example, you have a=3,b=10,c=7a=3, b=10, c=7 with λ+μ=10\lambda + \mu = 10 and λμ=21\lambda \mu = 21. Shouldn't take too long to notice that λ=3\lambda = 3 and μ=7\mu = 7 are valid conditions which make those properties true.

So then you have (x+33)(x+73)=0(x+\frac{3}{3})(x+\frac{7}{3})=0 which can be simplified further.

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