+0

# dont understand how to do this

0
628
3
+262

4<PQ<7

Aug 4, 2015

#2
+100256
+15

Hi Chris and Sabi,

I want to try this one too :)

The parabola at the top is concave up so it is the one with the positive leading coefficient.

so you want

$$\\4<(\frac{1}{2}x^2+7)-(\frac{-1}{4}x^2+3x)<7\\\\ 4<(\frac{2x^2}{4}+7)-(\frac{-x^2}{4}+3x)<7\\\\ 4<\frac{2x^2}{4}+7+\frac{x^2}{4}-3x<7\\\\ 4<\frac{3x^2}{4}+7-3x<7\\\\ -3<\frac{3x^2}{4}-3x<0\\\\ -12<3x^2-12x<0\\\\ -4 -4 x^2-4x>-4\qquad and \quad x(x-4)<0\\\\ x^2-4x+4>0\qquad and \quad x(x-4)<0\\\\ (x-2)^2>0\qquad and \quad x(x-4)<0\\\\$$

$$y=(x-2)^2$$    is a concave up parabola with a double root at x=2 so it is going to be positive for all valures of y except  x=2

$$y=x(x-4)$$  is a concave up parabola with roots at x=0 and x=4 . It will be BELOW the xaxis (negative) between the 2 roots.      Negative for   $$0 These will both be true when$$0      EXCEPT  when x=2

This is identical to CPhill's answer so by the law of averages we are probably both correct.   LOL

Aug 5, 2015

#1
+99702
+15

I possibly don't understand this, but I think you want to find the values where the differences in the height of the curves is more than 4 but less than 7....so we have....

[(1/2)x^2 + 7] - [(-1/4)x^2 + 3x] > 4   simplify

(3/4)x^2 - 3x + 7 > 4

(3/4)x^2 - 3x  + 3  > 0      and this is > 0 everywhere except  at x = 2 → ( -∞, 2) U (2, ∞)

See the graph, here....https://www.desmos.com/calculator/komufu5yah

And we also have

[(1/2)x^2 + 7] - [(-1/4)x^2 + 3x] < 7   simplify

(3/4)x^2 -3x  + 7 < 7

(3/4)x^2 - 3x < 0        and the interval where this happens is (0, 4)

See the graph here.....https://www.desmos.com/calculator/eepspvllj9

Now......only the intersection of these intervals will  satisfy this inequality.......these will be (0, 2) and (2, 4)

Note that when x = 2....the difference in the height of the curves is just 4......we need it larger than 4, though

See the graph here....https://www.desmos.com/calculator/hpb4r3uo1m

This one was a little tricky....I hope I haven't missed anything !!!

Aug 4, 2015
#2
+100256
+15

Hi Chris and Sabi,

I want to try this one too :)

The parabola at the top is concave up so it is the one with the positive leading coefficient.

so you want

$$\\4<(\frac{1}{2}x^2+7)-(\frac{-1}{4}x^2+3x)<7\\\\ 4<(\frac{2x^2}{4}+7)-(\frac{-x^2}{4}+3x)<7\\\\ 4<\frac{2x^2}{4}+7+\frac{x^2}{4}-3x<7\\\\ 4<\frac{3x^2}{4}+7-3x<7\\\\ -3<\frac{3x^2}{4}-3x<0\\\\ -12<3x^2-12x<0\\\\ -4 -4 x^2-4x>-4\qquad and \quad x(x-4)<0\\\\ x^2-4x+4>0\qquad and \quad x(x-4)<0\\\\ (x-2)^2>0\qquad and \quad x(x-4)<0\\\\$$

$$y=(x-2)^2$$    is a concave up parabola with a double root at x=2 so it is going to be positive for all valures of y except  x=2

$$y=x(x-4)$$  is a concave up parabola with roots at x=0 and x=4 . It will be BELOW the xaxis (negative) between the 2 roots.      Negative for   $$0 These will both be true when$$0      EXCEPT  when x=2

This is identical to CPhill's answer so by the law of averages we are probably both correct.   LOL

Melody Aug 5, 2015
#3
+99702
+5

Yeah....we couldn't both be wrong......LOL!!!!

Aug 5, 2015