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i cant understand how i got the math problem wrong
- Thread starter lil1ch1ck
- Start date
Hello, lil1ch1ck!
But 9 doesn't check, does it?
You have: \(\displaystyle \,\sqrt{2\cdot9\,-\,5}\,-\,\sqrt{9\,-\,3} \:=\:\sqrt{13}\,-\,\sqrt{6}\) . . . which does not equal 1.
You squared illegally, didn't you? *
Isolate a radical: \(\displaystyle \,\sqrt{2x\,-\,5}\;=\;\sqrt{x\,-\,3}\,+\,1\)
Square both sides: \(\displaystyle \,\left(\sqrt{2x\,-\,5}\right)^2\;=\;\left(\sqrt{x\,-\,3}\,+\,1\right)^2\)
And we get: \(\displaystyle \,2x\,-\,5\;=\;x\,-\,3\,+\,2\sqrt{x\,-\,3}\,+\,1\)
\(\displaystyle \;\;\)which simplifies to: \(\displaystyle \,x\,-\,3\;=\;2\sqrt{x\,-\,3}\)
Square both sides:\(\displaystyle \,(x\,-\,3)^2\;=\;\left(2\sqrt{x\,-\,3})^2\)
And we get: \(\displaystyle \,x^2\,-\,6x\,+\,9\;=\;4(x\,-\,3)\)
\(\displaystyle \;\;\)which simplifies to: \(\displaystyle \,x^2\,-\,10x\,+\,21\;=\;0\) . . . a quadratic
\(\displaystyle \;\;\)which factors: \(\displaystyle \,(x\,-\,3)(x\,-\,7)\;=\;0\)
\(\displaystyle \;\;\)and has roots: \(\displaystyle \,x\:=\;3,\;7\)
Since we squared the equation, we must check for extraneous roots.
And we find that both roots are acceptable.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
*
It is true that: \(\displaystyle \,5\,-\,2\;=\;3\)
\(\displaystyle \;\;\)But it is not true that: \(\displaystyle \,5^2\,-\,2^2\;=\;3^2\)
You cannot run through an expression and square the terms like that.
\(\displaystyle \sqrt{2x\,-\,5}\,-\,\sqrt{x\,-\,3}\;=\;1\)
Can you explain why the answer is 3 or 7?
When i solved it myself, i got 9 as an answer.
But 9 doesn't check, does it?
You have: \(\displaystyle \,\sqrt{2\cdot9\,-\,5}\,-\,\sqrt{9\,-\,3} \:=\:\sqrt{13}\,-\,\sqrt{6}\) . . . which does not equal 1.
You squared illegally, didn't you? *
Isolate a radical: \(\displaystyle \,\sqrt{2x\,-\,5}\;=\;\sqrt{x\,-\,3}\,+\,1\)
Square both sides: \(\displaystyle \,\left(\sqrt{2x\,-\,5}\right)^2\;=\;\left(\sqrt{x\,-\,3}\,+\,1\right)^2\)
And we get: \(\displaystyle \,2x\,-\,5\;=\;x\,-\,3\,+\,2\sqrt{x\,-\,3}\,+\,1\)
\(\displaystyle \;\;\)which simplifies to: \(\displaystyle \,x\,-\,3\;=\;2\sqrt{x\,-\,3}\)
Square both sides:\(\displaystyle \,(x\,-\,3)^2\;=\;\left(2\sqrt{x\,-\,3})^2\)
And we get: \(\displaystyle \,x^2\,-\,6x\,+\,9\;=\;4(x\,-\,3)\)
\(\displaystyle \;\;\)which simplifies to: \(\displaystyle \,x^2\,-\,10x\,+\,21\;=\;0\) . . . a quadratic
\(\displaystyle \;\;\)which factors: \(\displaystyle \,(x\,-\,3)(x\,-\,7)\;=\;0\)
\(\displaystyle \;\;\)and has roots: \(\displaystyle \,x\:=\;3,\;7\)
Since we squared the equation, we must check for extraneous roots.
And we find that both roots are acceptable.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
*
It is true that: \(\displaystyle \,5\,-\,2\;=\;3\)
\(\displaystyle \;\;\)But it is not true that: \(\displaystyle \,5^2\,-\,2^2\;=\;3^2\)
You cannot run through an expression and square the terms like that.
wow, thx
yeah sry bout my mistake. i have lots of trouble with the rules since they sometimes get jumbled up in my head. i understand my mistake now. earlier i had asked the same question on yahoo but their explanation was rather confusing & instead of using symbols, they used words. thanks for ur help & for using symbols instead of words. im new to this site so yeah
yeah sry bout my mistake. i have lots of trouble with the rules since they sometimes get jumbled up in my head. i understand my mistake now. earlier i had asked the same question on yahoo but their explanation was rather confusing & instead of using symbols, they used words. thanks for ur help & for using symbols instead of words. im new to this site so yeah