Advertisements
Advertisements
Question
Solve the following equation using the formula:
`2/3x = -1/6x^2 - 1/3`
Solution
`2/3x = -1/6x^2 - 1/3`
`\implies` 4x = – x2 – 2
`implies` x2 + 4x + 2 = 0
Here a = 1, b = 4 and c = 2
Then `x = (-b +- sqrt(b^2 - 4ac))/(2a)`
= `(-(-4) +- sqrt((4)^2-4(1)(2)))/(2(1))`
= `(-4 +- sqrt(8))/2`
= `(-4 +- 2sqrt(2))/2`
= `-2 +- sqrt(2)`
APPEARS IN
RELATED QUESTIONS
Without solving the following quadratic equation, find the value of ‘p' for which the given equation has real and equal roots:
x2 + (p – 3)x + p = 0.
Without solving, comment upon the nature of roots of the following equations
25x2 – 10x +1=0
Find the value of ‘p’, if the following quadratic equation have equal roots:
4x2 – (p – 2)x + 1 = 0
Solve `4/(x + 2) - 1/(x + 3) = 4/(2x + 1)`
Find the quadratic equation, whose solution set is:
{3, 5}
`2^2x-3.2^((x+2))+32=0`
Find the value of k for which the equation 3x2 – 6x + k = 0 has distinct and real roots.
Solve using the quadratic formula x² – 4x + 1 = 0
Check whether the following are quadratic equations: `sqrt(3)x^2 - 2x + (3)/(5) = 0`
Solve the following equation by using formula :
a (x2 + 1) = (a2+ 1) x , a ≠ 0
