Question :-
If the LCM of (x - a) and (x - b) is [tex]x^2 [/tex] - 2x + 1, then the values of a and b respectively are...
Step-by-step explanation:
Given that, LCM of (x - a) and (x - b) is [tex]x^2 [/tex] - 2x + 1
So,
[tex]\sf \: (x - a)(x - b) = {x}^{2} - 2x + 1 \\ \\ [/tex]
[tex]\sf \: x(x - b) - a(x - b) = {x}^{2} - 2x + 1 \\ \\ [/tex]
[tex]\sf \: {x}^{2} - bx - ax + ab = {x}^{2} - 2x + 1 \\ \\ [/tex]
[tex]\sf \: {x}^{2} - (b + a)x + ab = {x}^{2} - 2x + 1 \\ \\ [/tex]
On comparing, we get
[tex]\sf \: a + b = 2 \: \sf\implies \: b = 2 - a \\ \\ [/tex]
and
[tex]\sf \: ab = 1 \\ \\ [/tex]
On substituting the value of b, we get
[tex]\sf \: a(2 - a) = 1 \\ \\ [/tex]
[tex]\sf \: 2a - {a}^{2} = 1 \\ \\ [/tex]
[tex]\sf \: {a}^{2} - 2a + 1 = 0 \\ \\ [/tex]
[tex]\sf \: {a}^{2} - 2a + {1}^{2} = 0 \\ \\ [/tex]
[tex]\sf \: {a}^{2} - 2(a)(1) + {1}^{2} = 0 \\ \\ [/tex]
[tex]\sf \: {(a - 1)}^{2} = 0 \\ \\ [/tex]
[tex]\sf \: a - 1 = 0 \\ \\ [/tex]
[tex]\sf\implies \sf \: a = 1 \\ \\ [/tex]
[tex]\sf\implies \sf \: b = 1 \\ \\ [/tex]
Hence,
If the LCM of (x - a) and (x - b) is [tex]x^2 [/tex] - 2x + 1, then the values of a and b respectively are 1 and 1.
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]
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Answers & Comments
Question :-
If the LCM of (x - a) and (x - b) is [tex]x^2 [/tex] - 2x + 1, then the values of a and b respectively are...
Step-by-step explanation:
Given that, LCM of (x - a) and (x - b) is [tex]x^2 [/tex] - 2x + 1
So,
[tex]\sf \: (x - a)(x - b) = {x}^{2} - 2x + 1 \\ \\ [/tex]
[tex]\sf \: x(x - b) - a(x - b) = {x}^{2} - 2x + 1 \\ \\ [/tex]
[tex]\sf \: {x}^{2} - bx - ax + ab = {x}^{2} - 2x + 1 \\ \\ [/tex]
[tex]\sf \: {x}^{2} - (b + a)x + ab = {x}^{2} - 2x + 1 \\ \\ [/tex]
On comparing, we get
[tex]\sf \: a + b = 2 \: \sf\implies \: b = 2 - a \\ \\ [/tex]
and
[tex]\sf \: ab = 1 \\ \\ [/tex]
On substituting the value of b, we get
[tex]\sf \: a(2 - a) = 1 \\ \\ [/tex]
[tex]\sf \: 2a - {a}^{2} = 1 \\ \\ [/tex]
[tex]\sf \: {a}^{2} - 2a + 1 = 0 \\ \\ [/tex]
[tex]\sf \: {a}^{2} - 2a + {1}^{2} = 0 \\ \\ [/tex]
[tex]\sf \: {a}^{2} - 2(a)(1) + {1}^{2} = 0 \\ \\ [/tex]
[tex]\sf \: {(a - 1)}^{2} = 0 \\ \\ [/tex]
[tex]\sf \: a - 1 = 0 \\ \\ [/tex]
[tex]\sf\implies \sf \: a = 1 \\ \\ [/tex]
So,
[tex]\sf\implies \sf \: b = 1 \\ \\ [/tex]
Hence,
If the LCM of (x - a) and (x - b) is [tex]x^2 [/tex] - 2x + 1, then the values of a and b respectively are 1 and 1.
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]