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Mathematics 14 Online
OpenStudy (precal):

Solving an inequality x-7 lessthan or equal to square root of (x-5)

OpenStudy (precal):

sorry I am using my ipad unable to use equation editor and I don't know latex

OpenStudy (phi):

\[ x-7 \le \sqrt{x-5 } \] start by squaring both sides

OpenStudy (precal):

yes and I got x^2-14x+49 > x-5 sorry can't not able to type in greater than or equal to

OpenStudy (precal):

x^2-15x+54>0

OpenStudy (phi):

originally you has less than or equal. which is it ?

OpenStudy (precal):

I know the solution is x> 0 sorry meant greater than

OpenStudy (precal):

do I factor that or and by the way I meant the solution is x >9

OpenStudy (phi):

yes factor

OpenStudy (precal):

sorry I am on my ipad not my computer am I suppose to factor the quadratic?

OpenStudy (phi):

factor x^2-15x+54>= 0

OpenStudy (precal):

how does x>6 play into the solution? I mean x>9 would include that interval

OpenStudy (phi):

you get (x-6)(x-9) > 0 if you think A*B>0 that means both A and B are positive or both A and B are negative. so x>6 and x>9 (at the same time) or x<6 and x<9 (at the same time) ** this solution gives negative square roots so ignore it for the first solution, for x to be bigger than 6 and bigger than 9 means just x>=9

OpenStudy (phi):

to review: (x-6) * (x-9) will be a negative number if x is (for example) 7: (7-6) * (7-9)= 1 * -2 = -2 and we want (x-6) * (x-9) >0 so if we need both x>6 and x>9, that simplifies to x>9

OpenStudy (precal):

thanks

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