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3 years ago

Help with homework plzzzzz!!!!!!!!!!!!!

Mathematics

1 answer:

Zanzabum3 years ago

7 0

I'm not good with intigers, but if you added two negative intigers together wouldn't it still be negative? If you added -1 to -2 you'd get -3. It would always be less than the two original integers.

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Vladimir79 [104]

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soldier1979 [14.2K]

Answer:

Step-by-step explanation:

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4 years ago

How many extraneous solutions does the equation below have?

aev [14]

Answer:

A solution is said to be extraneous, if it is a zero of the equation, but it does not satisfy the equation,when substituted in the original equation,L.H.S≠R.H.S.

The given equation consisting of variable , m is

$\frac{2 m}{2 m+3} -\frac{2 m}{2 m-3}=1\\\\ 2 m[\frac{1}{2 m+3} -\frac{1}{2 m-3}]=1\\\\ 2 m\times \frac{[2 m-3 -2 m- 3]}{4m^2-9}=1\\\\ -6 \times 2 m=4 m^2 -9\\\\ 4 m^2 +1 2 m -9=0\\\\m=\frac{-12 \pm\sqrt{12^2-4 \times 4 \times (-9)}}{2\times 4}\\\\m=\frac{-12 \pm \sqrt {144+144}}{8}\\\\m=\frac{-12 \pm \sqrt {288}}{8}\\\\m=\frac{-12 \pm 12 \sqrt{2}}{8}\\\\m=\frac{3}{2}\times(-1 \pm \sqrt{2})$

None of the two solution

$m=\frac{3}{2}\times(-1 +\sqrt{2}) and , m=\frac{3}{2}\times(-1 -\sqrt{2})$ , is extraneous.

Here, L.H.S= R.H.S

Option A: 0→ extraneous

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