Someone help me!!
square root = sqrt
(5x^2-5x-10)sqrt(x+7) + (2x+6)sqrt(x+2) >= x^3+13x^2-6x-28
2 answers
"12th grade" tells us nothing about what you need help with. Is this trig, calc, algebra, or what SUBJECT? I tutor in English and social studies and communication and journalism. I can't help you. The proper tutor may never respond to "12th grade'.
(5x^2-5x-10)√(x+7) + (2x+6)√(x+2) ≥ x^3+13x^2-6x-28 , where x ≥ -2
A quick mental inspection shows that when x = 2, the roots become exact.
Could it be ? could it be?
testing for x = 2
LS = (20-10-10)√9 + (4+6)√4
= 0 + 20
= 20
RS = 8 + 4(13) - 12 - 28
= 20
so x = 2 , at the equality
Well, that was a "lucky" but logical guess .
I graphed the LS and the RS as separated functions to see where they are relative to teach other:
http://www.wolframalpha.com/input/?i=plot+y+%3D+(5x%5E2-5x-10)%E2%88%9A(x%2B7)+%2B+(2x%2B6)%E2%88%9A(x%2B2),+y++%3D+x%5E3%2B13x%5E2-6x-28+for+x+%3D+-2+to+5
testing at x = 0
LS = -10√7 + 6√2 = appr -17.98
RS = -28
So LS ≥RS at x = 0
It is highly unlikely, looking at my graphs, that there are any other solutions, so
-2 ≤ x ≤ 2
A quick mental inspection shows that when x = 2, the roots become exact.
Could it be ? could it be?
testing for x = 2
LS = (20-10-10)√9 + (4+6)√4
= 0 + 20
= 20
RS = 8 + 4(13) - 12 - 28
= 20
so x = 2 , at the equality
Well, that was a "lucky" but logical guess .
I graphed the LS and the RS as separated functions to see where they are relative to teach other:
http://www.wolframalpha.com/input/?i=plot+y+%3D+(5x%5E2-5x-10)%E2%88%9A(x%2B7)+%2B+(2x%2B6)%E2%88%9A(x%2B2),+y++%3D+x%5E3%2B13x%5E2-6x-28+for+x+%3D+-2+to+5
testing at x = 0
LS = -10√7 + 6√2 = appr -17.98
RS = -28
So LS ≥RS at x = 0
It is highly unlikely, looking at my graphs, that there are any other solutions, so
-2 ≤ x ≤ 2
1 answer