If x^2 + 4x - 1= 0, find x. Value of x is+ve...
Answers
Answer:
I think in question x² + 4x - 1 = 0 is wrong
the correct equation is x² + 4x + 1 = 0
\begin{gathered} {x}^{2} + 4x + 1 = 0 \\ {x}^{2} + 4x = - 1 \\ x(x + 4) = - 1 \\ x = \frac{ - 1}{x + 4} \\ \\ x + \frac{1}{x} = \frac{ - 1}{x + 4} + \frac{1}{ \frac{ - 1}{x + 4} } \\ x + \frac{1}{x} = - 1( \frac{1}{x + 4} + (x + 4)) \\ x + \frac{1}{x} = - 1 (\frac{ 1 + (x + 4)(x + 4)}{(x + 4)} \\ x + \frac{1}{x} = - 1( \frac{1 + {x}^{2} + 8x + 16 }{x + 4}) \\ x + \frac{1}{x} = - 1( \frac{ {x}^{2} + 8x + 17 }{x + 4} ) \\ x + \frac{1}{x} = - 1( \frac{( {x}^{2} + 4x + 1) + 4x + 16}{x + 4}) \\ x + \frac{1}{x} = - 1( \frac{(0) + 4x + 16}{x + 4} ) \\ x + \frac{1}{x} = - 1(\frac{4(x + 4)}{x + 4} ) \\ x + \frac{1}{x} = - 4\end{gathered}
x
2
+4x+1=0
x
2
+4x=−1
x(x+4)=−1
x=
x+4
−1
x+
x
1
=
x+4
−1
+
x+4
−1
1
x+
x
1
=−1(
x+4
1
+(x+4))
x+
x
1
=−1(
(x+4)
1+(x+4)(x+4)
x+
x
1
=−1(
x+4
1+x
2
+8x+16
)
x+
x
1
=−1(
x+4
x
2
+8x+17
)
x+
x
1
=−1(
x+4
(x
2
+4x+1)+4x+16
)
x+
x
1
=−1(
x+4
(0)+4x+16
)
x+
x
1
=−1(
x+4
4(x+4)
)
x+
x
1
=−4