Solve the equation: x^2 + 7x − 3 = 0
Is this a parabola?
Looks like a quadratic.
its a quadratic
There are two ways to solve. By factoring and by using the quadratic equation.
use the formula to find it x={-b+-sqrt(b^2-4ac)}/2a where a=1, b=7 and c=(-3)
use Quadratic for this.
In this case, you sue the quadratic formula: \[\huge x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
you use*
Where: a represents the coefficient of x^2 b represents the coefficient of x c represents the constant
so...its -7+- sqr root of 61 all over 2?...
According to me this question can be done in much easier way rather than using the quadratic formula so what u can do according to me is as follows; \[x(x+7)=3\] therfore x can either be 3 or -7
@u0860867 That goes against everything about mathematics. Those kind of shortcuts are the ones that get you 0 marks. I see people who do that, and teachers go beserk if you ever do that.
And you're correct @brandon57 Well Done!
@Azteck Sorry for sounding a bit rude but i don't get what's wrong with the way i worked out the solution pls can u explain me what wrong with my working out since i have a similar question i need help with.
When solving quadratic equations, you must let one side equal to ZERO!!! You CANNOT make a side equal to 3 or any other number. IT MUST BE ZERO!
You will always be wrong if you do not let one side equal to zero when solving for quadratics. ANd questions like yours @u0860867 would probably not include quadratic equations rather than linear algebra.
@Azteck - thanks i get what u are talking about now i think that is the reason i am not getting the right answer .
If you weren't getting the right answer, you should not tell others to do that same thing you did that got you the wrong answer. Please do not sabotage others for their own benefit. Thank you.
@Azteck it wasn't exactly the same question and the method i mentioned above does work at times and the other questions that i did get right i used the same working out so i am just mentioning my opinion of doing the question i might not have got the whole question right but that does not mean i want to sabotage someone else's question for my benefit.
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