if a = 4-√15 , find
(i) a² + 1/a²
(ii) a³ + 1/a³
(iii) a⁴ + 1/a⁴
Answers
Answer:
Given : In a two-digit number, the sum of the digits is 5 more than the units digit. The difference between the original number and the sum of digits is 10 more than the number formed by reversing the digits. Then find the difference between the digits.
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\begin{gathered} \frak{Let \: us \: assume} \rightarrow \begin{cases} \frak{The \: units \: digits \: be \: \green{a}} \\ \\ \frak{The \: tens \: digit \: be \: \green{b}}\end{cases}\end{gathered}
Letusassume→
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⎪
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⎪
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Theunitsdigitsbea
Thetensdigitbeb
Hence, we get
\begin{gathered} \implies \frak{ a + b = 5 + b} \\ \\ \implies \frak{ \pink{a = 5}}\end{gathered}
⟹a+b=5+b
⟹a=5
After which the original number is
\dashrightarrow \frak{ \frak{Original \: Number : \underline{10a + b}}}⇢OriginalNumber:
10a+b
Reversing the digit we will get the new number
\dashrightarrow \frak{New \: Number = \underline{10b + a}}⇢NewNumber=
10b+a
Now, according to the question putting all the required values to get the equation
\begin{gathered} \rightharpoondown \frak{10a + b - (a + b) = 10 + 10b + a} \\ \\ \rightharpoonup \frak{10a + b - a - b = 10 + 10b + a} \\ \\ \rightharpoondown \frak{9a + \cancel b - \cancel b = 10 + 10b + a} \\ \\ \rightharpoonup \frak{9a - a = 10 + 10b} \\ \\ \rightharpoondown \frak{8(5) - 10 = 10b} \\ \\ \rightharpoonup \frak{40 - 10 = 10b} \\ \\ \rightharpoondown \frak{10b = 30} \\ \\ \star \quad \underline{ \boxed{ \frak{ \purple{b = 3}}}} \end{gathered}
⇁10a+b−(a+b)=10+10b+a
⇀10a+b−a−b=10+10b+a
⇁9a+
b
−
b
=10+10b+a
⇀9a−a=10+10b
⇁8(5)−10=10b
⇀40−10=10b
⇁10b=30
⋆
b=3
\dag \: \: \underline{ \frak{As \: we \: know \: that} }:†
Asweknowthat
:
\begin{gathered} \frak{ \blue{a = 5 } \: and\: \pink{b = 3}} \\ \\ \therefore \sf Difference \: between \: the \: numbers = \frak{5 - 3 =\red{2}}\end{gathered}
a=5andb=3
∴Differencebetweenthenumbers=5−3=2