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secondary 3 | A Maths
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hi need help thanku!
Date Posted: 2 years ago
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x² - 2x = a
a = √ 2a+3
square at both sides
a² = 2a+3
a² - 2a - 3 = 0
(a-3)⋅(a+1) = 0
a₁ = 3 = x² - 2x → ①
a₂ = -1 = x² - 2x → ②
① → x² - 2x - 3 = 0
(x-3)⋅(x+1) = 0
x₁ = 3
x₂ = -1
② → x² - 2x + 1 = 0
(x-1)² = 0
x₃ = 1
a = √ 2a+3
square at both sides
a² = 2a+3
a² - 2a - 3 = 0
(a-3)⋅(a+1) = 0
a₁ = 3 = x² - 2x → ①
a₂ = -1 = x² - 2x → ②
① → x² - 2x - 3 = 0
(x-3)⋅(x+1) = 0
x₁ = 3
x₂ = -1
② → x² - 2x + 1 = 0
(x-1)² = 0
x₃ = 1
Some algebraic processes used in solving equations, may result in "extraneous" solutions, which means solutions that are not actually valid solutions, while some algebraic processes may result in "missing" solutions, which means less solutions than there actually are.
Squaring both sides is a right method here, but need to be aware that squaring both sides is one process which may result in extraneous solutions, so after finding the solutions, we need to verify the validity of each solution.
By substituting x=3 & x=-1 into the original equation, we find that they are both valid solutions. However, substituting x=1, we get LHS=-1, but with just the square root sign, RHS should be interpreted as +1, therefore x=1 need to be rejected.
For more information about "extraneous" solutions and "missing" solutions, can check https://www.google.com/search?q=extraneous+solutions+wiki&sxsrf=AOaemvKYSKw-ZgkoLll7xmecjs5RU-Wt6g%3A1639578916980&source=hp&ei=JP25YY_7OI_ez7sP7YWOyAQ&iflsig=ALs-wAMAAAAAYboLNJO1ZHc6lmHcWMaLbBHJuHLJGFJt&oq=extrane&gs_lcp=Cgdnd3Mtd2l6EAEYADIECCMQJzIECCMQJzIFCAAQgAQyBQgAEIAEMgUIABCABDIFCAAQgAQyBQgAEIAEMgUIABCABDIFCAAQgAQyBQgAEIAEOgcIIxDqAhAnOgsILhCABBCxAxCDAToICC4QsQMQgwE6CwgAEIAEELEDEIMBOggILhCABBCxAzoICAAQsQMQgwE6CAgAEIAEELEDOg4ILhCABBCxAxDHARDRAzoFCAAQsQM6CwguEIAEEMcBEK8BOgUILhCABFCVGljxKmCkPmgCcAB4AIAB5wGIAd0FkgEFNS4xLjGYAQCgAQGwAQo&sclient=gws-wiz. If the link does not work, google "extraneous solutions wikipedia".
Squaring both sides is a right method here, but need to be aware that squaring both sides is one process which may result in extraneous solutions, so after finding the solutions, we need to verify the validity of each solution.
By substituting x=3 & x=-1 into the original equation, we find that they are both valid solutions. However, substituting x=1, we get LHS=-1, but with just the square root sign, RHS should be interpreted as +1, therefore x=1 need to be rejected.
For more information about "extraneous" solutions and "missing" solutions, can check https://www.google.com/search?q=extraneous+solutions+wiki&sxsrf=AOaemvKYSKw-ZgkoLll7xmecjs5RU-Wt6g%3A1639578916980&source=hp&ei=JP25YY_7OI_ez7sP7YWOyAQ&iflsig=ALs-wAMAAAAAYboLNJO1ZHc6lmHcWMaLbBHJuHLJGFJt&oq=extrane&gs_lcp=Cgdnd3Mtd2l6EAEYADIECCMQJzIECCMQJzIFCAAQgAQyBQgAEIAEMgUIABCABDIFCAAQgAQyBQgAEIAEMgUIABCABDIFCAAQgAQyBQgAEIAEOgcIIxDqAhAnOgsILhCABBCxAxCDAToICC4QsQMQgwE6CwgAEIAEELEDEIMBOggILhCABBCxAzoICAAQsQMQgwE6CAgAEIAEELEDOg4ILhCABBCxAxDHARDRAzoFCAAQsQM6CwguEIAEEMcBEK8BOgUILhCABFCVGljxKmCkPmgCcAB4AIAB5wGIAd0FkgEFNS4xLjGYAQCgAQGwAQo&sclient=gws-wiz. If the link does not work, google "extraneous solutions wikipedia".
to Shyann,
please google "square root symbol wikipedia", to read about how to interpret square root symbol when there is no + or - in front of it.
admittedly, there are differing views on this, so it is better that you verify with your school maths teacher, on what is correct interpretation for singapore maths syllabus.
please google "square root symbol wikipedia", to read about how to interpret square root symbol when there is no + or - in front of it.
admittedly, there are differing views on this, so it is better that you verify with your school maths teacher, on what is correct interpretation for singapore maths syllabus.
See 1 Answer
Some algebraic processes used in solving equations, may result in "extraneous" solutions, which means solutions that are not actually valid solutions, while some algebraic processes may result in "missing" solutions, which means less solutions than there actually are.
Squaring both sides is a right method here, but need to be aware that squaring both sides is one process which may result in extraneous solutions, so after finding the solutions, we need to verify the validity of each solution.
By substituting x=3 & x=-1 into the original equation, we find that they are both valid solutions. However, substituting x=1, we get LHS=-1, but with just the square root sign, RHS should be interpreted as +1, therefore x=1 need to be rejected.
For more information about "extraneous" solutions and "missing" solutions, can check https://www.google.com/search?q=extraneous+solutions+wiki&sxsrf=AOaemvKYSKw-ZgkoLll7xmecjs5RU-Wt6g%3A1639578916980&source=hp&ei=JP25YY_7OI_ez7sP7YWOyAQ&iflsig=ALs-wAMAAAAAYboLNJO1ZHc6lmHcWMaLbBHJuHLJGFJt&oq=extrane&gs_lcp=Cgdnd3Mtd2l6EAEYADIECCMQJzIECCMQJzIFCAAQgAQyBQgAEIAEMgUIABCABDIFCAAQgAQyBQgAEIAEMgUIABCABDIFCAAQgAQyBQgAEIAEOgcIIxDqAhAnOgsILhCABBCxAxCDAToICC4QsQMQgwE6CwgAEIAEELEDEIMBOggILhCABBCxAzoICAAQsQMQgwE6CAgAEIAEELEDOg4ILhCABBCxAxDHARDRAzoFCAAQsQM6CwguEIAEEMcBEK8BOgUILhCABFCVGljxKmCkPmgCcAB4AIAB5wGIAd0FkgEFNS4xLjGYAQCgAQGwAQo&sclient=gws-wiz. If the link does not work, google "extraneous solutions wikipedia".
Squaring both sides is a right method here, but need to be aware that squaring both sides is one process which may result in extraneous solutions, so after finding the solutions, we need to verify the validity of each solution.
By substituting x=3 & x=-1 into the original equation, we find that they are both valid solutions. However, substituting x=1, we get LHS=-1, but with just the square root sign, RHS should be interpreted as +1, therefore x=1 need to be rejected.
For more information about "extraneous" solutions and "missing" solutions, can check https://www.google.com/search?q=extraneous+solutions+wiki&sxsrf=AOaemvKYSKw-ZgkoLll7xmecjs5RU-Wt6g%3A1639578916980&source=hp&ei=JP25YY_7OI_ez7sP7YWOyAQ&iflsig=ALs-wAMAAAAAYboLNJO1ZHc6lmHcWMaLbBHJuHLJGFJt&oq=extrane&gs_lcp=Cgdnd3Mtd2l6EAEYADIECCMQJzIECCMQJzIFCAAQgAQyBQgAEIAEMgUIABCABDIFCAAQgAQyBQgAEIAEMgUIABCABDIFCAAQgAQyBQgAEIAEOgcIIxDqAhAnOgsILhCABBCxAxCDAToICC4QsQMQgwE6CwgAEIAEELEDEIMBOggILhCABBCxAzoICAAQsQMQgwE6CAgAEIAEELEDOg4ILhCABBCxAxDHARDRAzoFCAAQsQM6CwguEIAEEMcBEK8BOgUILhCABFCVGljxKmCkPmgCcAB4AIAB5wGIAd0FkgEFNS4xLjGYAQCgAQGwAQo&sclient=gws-wiz. If the link does not work, google "extraneous solutions wikipedia".
√1=±1
to Shyann,
please google "square root symbol wikipedia", to read about how to interpret square root symbol when there is no + or - in front of it.
admittedly, there are differing views on this, so it is better that you verify with your school maths teacher, on what is correct interpretation for singapore maths syllabus.
please google "square root symbol wikipedia", to read about how to interpret square root symbol when there is no + or - in front of it.
admittedly, there are differing views on this, so it is better that you verify with your school maths teacher, on what is correct interpretation for singapore maths syllabus.
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