Which interval for the graphed function contains the local minimum of 0
2 answers:
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<span><u><em>The correct answer is: </em></u>
C) his solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.
<u><em>Explanation: </em></u>
<span><u>To solve this rational equation, we cross multiply: </u>
8*(x-4)=2*(x+2).
<u>Using the distributive property, we have </u>
8*x-8*4=2*x+2*2;
8x-32=2x+4.
<u>Subtract 2x from each side: </u>
8x-32-2x=2x+4-2x;
6x-32=4.
<u>Add 32 to each side: </u>
6x-32+32=4+32;
6x=36.
<u>Divide both sides by 6:</u>
</span></span>
<span><span>=</span></span>
<span><span>;
x=6.
<u>Extraneous solutions</u> are solutions that come about in the problem but are not valid solutions to the problem; the only values of x that would give this sort of answer are -2 and 4, since these are the two values that would make one of the denominators 0 (we cannot divide by 0).</span></span>
C) his solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.
<u><em>Explanation: </em></u>
<span><u>To solve this rational equation, we cross multiply: </u>
8*(x-4)=2*(x+2).
<u>Using the distributive property, we have </u>
8*x-8*4=2*x+2*2;
8x-32=2x+4.
<u>Subtract 2x from each side: </u>
8x-32-2x=2x+4-2x;
6x-32=4.
<u>Add 32 to each side: </u>
6x-32+32=4+32;
6x=36.
<u>Divide both sides by 6:</u>
</span></span>
x=6.
<u>Extraneous solutions</u> are solutions that come about in the problem but are not valid solutions to the problem; the only values of x that would give this sort of answer are -2 and 4, since these are the two values that would make one of the denominators 0 (we cannot divide by 0).</span></span>