find the value of P and Q in
478
1P8
+26Q
=9Q9
Answers
Answered by
0
Answer:
Value of x+\frac{1}{x}=±\sqrt{29}x+
x
1
=±
29
Explanation:
Given x^{2}+\frac{1}{x^{2}}=27x
2
+
x
2
1
=27
\begin{gathered}\implies x^{2}+\frac{1}{x^{2}}+2\times x \times \frac{1}{x}\\=27+2\times x \times \frac{1}{x}\end{gathered}
⟹x
2
+
x
2
1
+2×x×
x
1
=27+2×x×
x
1
\begin{gathered}\implies x^{2}+\frac{1}{x^{2}}+2\times x \times \frac{1}{x}\\=27+2\end{gathered}
⟹x
2
+
x
2
1
+2×x×
x
1
=27+2
\implies \left(x+\frac{1}{x}\right)^{2}=29⟹(x+
x
1
)
2
=29
___________________________
By algebraic identity:
\boxed {a^{2}+2ab+b^{2}=(a+b)^{2}}
a
2
+2ab+b
2
=(a+b)
2
___________________________
\implies x+\frac{1}{x}=±\sqrt{29}⟹x+
x
1
=±
29
Therefore,
Value of x+\frac{1}{x}=±\sqrt{29}x+
x
1
=±
29
•••••
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