@ganeshie8
Select one: a. There will be 1 finite answer b. There will be 2 finite answers c. The only answer is zero d. No solution
i think no solution @prowrestler
is this correct @ganeshie8
that may be correct/incorrect, don't you want to know when an equation will have or not have solutions ?
what does an equation represent anyways ? \[\large \sqrt{x} +2 = \sqrt{x} + 3\] the left side exactly balances the right side ?
can you find a value of \(\large x\) such that the above equation holds ?
sorry to ask too many questions, but u seem to be only interested in answers and not in learning >.<
im interested in learning
i just was told the answer quickly and i wondered if it was right or not
so i was asking you if D. no solution was correct
One way to work this by using equality properties : \[\large \sqrt{x} +2 = \sqrt{x} + 3\] you're allowed to add/subtract a same number both side, so subtracting \(\large \sqrt{x}\) both sides, you get : \[\large \sqrt{x} +2-\color{red}{\sqrt{x}} = \sqrt{x} + 3 - \color{red}{\sqrt{x}}\] combine like terms and simplify, what do u get ?
no solution?
@prowrestler it means not equal acc. to your answer no solution
forget about options for now, you won't loose anything if u spend 10 minutes working this with me without rushing :o
so how am i going to finish my work
\[\large \sqrt{x} +2-\color{red}{\sqrt{x}} = \sqrt{x} + 3 - \color{red}{\sqrt{x}}\] combine like terms, what do u get ?
im not sure how o combine like terms for this question
you get : \[\large \sqrt{x} -\color{red}{\sqrt{x}} +2= \sqrt{x} - \color{red}{\sqrt{x}} + 3\] simplify : \[\large 0+2= 0+ 3\] \[\large 2= 3\]
is it possible 2 to equal 3 ever ?
no
so can we say there are NO solutionsfor the given equations ?
yes
no u can't say bcoz i think @ganeshie8 know that are unequal but has solution,...isn't it @ganeshie8
thats it! next time do not ask for opinions of others, you should figure these out on ur own and convince urself... it is okay even if you are wrong... most important thing is you need to convince urself and never use opinions of others as ur guide in math (just saying )
ok then can you HELP me on others
sure, i can help u work these on your own.. these are tricky when u work first time, but you would understand these after doing 1-2 problems
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