Advertisements
Advertisements
प्रश्न
Solve the following quadratic equations by factorization:
`(1 + 1/(x + 1))(1 - 1/(x - 1)) = 7/8`
Advertisements
उत्तर
`(1+1/(x+1))(1-1/(x-1))=7/8`
⇒ ` ((x + 1 + 1)/(x + 1))((x - 1 - 1)/(x - 1)) = 7/8`
⇒ `((x + 2)/(x + 1))((x - 2)/(x - 1)) = 7/8`
⇒ `(x^2 - 4)/(x^2 - 1) = 7/8`
⇒ 8x2 – 32 = 7x2 – 7
⇒ x2 = 25
⇒ x = ±5
APPEARS IN
संबंधित प्रश्न
Solve the following quadratic equations by factorization:
\[\frac{3}{x + 1} + \frac{4}{x - 1} = \frac{29}{4x - 1}; x \neq 1, -1, \frac{1}{4}\]
Solve the following quadratic equations by factorization:
\[3\left( \frac{3x - 1}{2x + 3} \right) - 2\left( \frac{2x + 3}{3x - 1} \right) = 5; x \neq \frac{1}{3}, - \frac{3}{2}\]
Find the values of k for which the quadratic equation \[\left( 3k + 1 \right) x^2 + 2\left( k + 1 \right)x + 1 = 0\] has equal roots. Also, find the roots.
If \[x = - \frac{1}{2}\],is a solution of the quadratic equation \[3 x^2 + 2kx - 3 = 0\] ,find the value of k.
Solve the following equation: `"a"("x"^2 + 1) - x("a"^2 + 1) = 0`
Solve the following quadratic equation by factorisation:
(2x + 3) (3x - 7) = 0
Solve the quadratic equation by factorisation method:
x2 – 15x + 54 = 0
A dealer sells a toy for ₹ 24 and gains as much percent as the cost price of the toy. Find the cost price of the toy.
A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.
The roots of the equation x2 + 3x – 10 = 0 are ______.
