Advertisements
Advertisements
Question
Solve:
`2(x^2 + 1/x^2) - (x + 1/x) = 11`
Solution
`2(x^2 + 1/x^2) - (x + 1/x) = 11`
Let `x + 1/x = y`
Squaring on both sides
`x^2 + 1/x^2 = y^2 - 2`
Putting these values in the given equation
2(y2 – 2) – y = 11
`=>` 2y2 – 4 – y – 11 = 0
`=>` 2y2 – y – 15 = 0
`=>` 2y2 – 6y + 5y – 15 = 0
`=>` 2y(y – 3) + 5(y – 3) = 0
`=>` (y – 3)(2y + 5) = 0
If y – 3 = 0 or 2y + 5 = 0
Then y = 3 or y =` (-5)/2`
`=> x + 1/x = 3` or `x + 1/x = (-5)/2`
`=> (x^2 + 1)/x = 3` or `(x^2 + 1)/x = (-5)/2`
`=> x^2 - 3x+ 1 = 0` or `2x^2 + 5x + 2 = 0`
`=> x = (-3 +- sqrt((-3)^2 - 4(1)(1)))/(2(1))` or `2x^2 + 4x + x + 2 = 0`
`=> x = (-3 +- sqrt5)/2` or `2x(x + 2) + 1(x + 2) = 0`
Then x = –2 and x = `(-1)/2`
APPEARS IN
RELATED QUESTIONS
Solve the equation 4x2 – 5x – 3 = 0 and give your answer correct to two decimal places
State whether the given equation is quadratic or not. Give reason.
`5/4m^2 - 7 = 0`
Solve `9/2 x = 5 + x^2`
Solve the following equation for x and give, in the following case, your answer correct to 2 decimal places:
`4x + 6/x + 13 = 0`
Solve: (2x+3)2 = 81
Find the value of x, if a + 1 = 0 and x2 + ax – 6 = 0
`sqrt7x^2-6x-13sqrt7=0`
`2x^2-x+1/8=0`
Choose the correct answer from the given four options :
If `(1)/(2)` is a root of the quadratic equation 4x2 – 4kx + k + 5 = 0, then the value of k is
Which of the following is not a quadratic equation?
