Advertisements
Advertisements
प्रश्न
Find the two consecutive positive even integers whose product is 288.
उत्तर
Let the two consecutive positive even integers be x and(x+2)
According to the given condition,
`x(x+2)=288`
⇒`x^2+2x-288=0`
⇒`x^2+18x-16x-288=0`
⇒`x(x+18)-16(x+18)=0`
⇒`(x+18)(x-16)=0`
⇒`x+18=0 or x-16=0`
⇒`x=-18 or x=16`
`∴x=16 ` (x is a positive even integer)
When` x=16 `
`x+2=16+2=18`
Hence, the required integers are 16 and 18.
APPEARS IN
संबंधित प्रश्न
Solve for x
`(x - 1)/(2x + 1) + (2x + 1)/(x - 1) = 2, "where x" != -1/2, 1`
Solve the following quadratic equations by factorization:
(x − 4) (x + 2) = 0
Solve the following quadratic equations by factorization:
`2/2^2-5/x+2=0`
The perimeter of a rectangular field is 82 m and its area is 400 m2. Find the breadth of the rectangle.
Factorise : m2 + 5m + 6.
Is there any real value of 'a' for which the equation x2 + 2x + (a2 + 1) = 0 has real roots?
Solve the following equation :
`("x" - 1)/("x" - 2) + ("x" - 3)/("x" - 4) = 3 1/3`
A two digit number is 4 times the sum of its digit and twice the product of its digit. Find the number.
Solve the following quadratic equation by factorisation method:
`x/(x + 1) + (x + 1)/x = (34)/(15') x ≠ 0, x ≠ -1`
Solve the equation:
`6(x^2 + (1)/x^2) -25 (x - 1/x) + 12 = 0`.