Solving quadratic inequalities algebraically without graphing/common-sensing the equality

ICESSRAIPCELOtblt

ICESSRAIPCELOtblt

Answered question

2022-04-03

Solving quadratic inequalities algebraically without graphing/common-sensing the equality signs
I've just been presented an inequality at GCSE (high school) level that, solving algebraically (the method used for linear inequalities), doesn't work correctly which has caught me by surprise. Research always leads to graphed or common-sensed solutions. Can anyone explain why this doesn't work and if there's an algebraic manipulation that leads to the right answers?
x249>0
Using the quadratic identity
(x+7)(x7)>0
Divide both sides by (x+7)...
x7>0
x>7
Divide both sides by (x7)...
x+7>0
x7
Alternatively,
x249>0
x2>49
x>±7
The negative solution has to be
x<7
but I don't know how that's arrived at through algebra.

Answer & Explanation

Marcos Boyer

Marcos Boyer

Beginner2022-04-04Added 12 answers

It is much fun and pretty enlightening to do this kind of exercises with a ping-pong between algebra and geometry. Anyway, if you want to avoid that , then we have the following algebraic facts which, funny enough, are pretty easy to prove precisely by means of geometry...but not only, as they follow from the definition of absolute value:
For For 0<a ,  {|x|a  axa |x|>a  x<a  or  x>a
Thus, in your case, we could do as follows (all algebra, not geometry):
x249>0  x2>49  taking square roots|x|>7  x<7   or   x>7
and in intervals notation, the solution is ;x(,7)(7,);.

Esteban Sloan

Esteban Sloan

Beginner2022-04-05Added 21 answers

If you want a purely algebraic solution then note that the inequality ab>0 is only satisfied when a>0 and b>0, or when a<0 and b<0. So in your example,
(x+7)(x7)>0,,
we need both x+7 and x7 to be greater than 0 (which gives us the solution x>7), or we need x+7 and x7 to both be less than 0 (which gives us the solution x<7). However, I think you are much less prone to error if you solve this geometrically.

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