Advertisements
Advertisements
प्रश्न
Solve the following equation: a2b2x2 + b2x - a2x - 1 = 0
उत्तर
a2b2x2 + b2x - a2x - 1 = 0
b2 x(a2 x + 1)- 1(a2 x + 1) = 0
(a2 x + 1)(b2 x- 1) = 0
x = `-1/"a"^2` , x = `1/"b"^2`
APPEARS IN
संबंधित प्रश्न
Solve for x
`(2x)/(x-3)+1/(2x+3)+(3x+9)/((x-3)(2x+3)) = 0, x!=3,`
Solve for x : `(x+1)/(x-1)+(x-1)/(x+2)=4-(2x+3)/(x-2);x!=1,-2,2`
Solve the equation `4/x-3=5/(2x+3); xne0,-3/2` for x .
The product of two successive integral multiples of 5 is 300. Determine the multiples.
Solve the following quadratic equations by factorization:
\[\frac{4}{x} - 3 = \frac{5}{2x + 3}, x \neq 0, - \frac{3}{2}\]
Solve the following quadratic equations by factorization: \[\frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3x - 1}, x \neq - 1, \frac{1}{3}\]
Find the value of k for which the following equations have real and equal roots:
\[\left( k + 1 \right) x^2 - 2\left( k - 1 \right)x + 1 = 0\]
Write the condition to be satisfied for which equations ax2 + 2bx + c = 0 and \[b x^2 - 2\sqrt{ac}x + b = 0\] have equal roots.
Solve the following equation: `"m"/"n" "x"^2 + "n"/"m" = 1- 2"x"`
The sum of the squares of three consecutive natural numbers is 110. Determine the numbers.
