Advertisements
Advertisements
प्रश्न
Two number differ by 4 and their product is 192. Find the numbers?
उत्तर
Let two required numbers be x and (x + 4)
Then according to question
x(x + 4) = 192
x2 + 4x - 192 = 0
x2 + 16x - 12x - 192 = 0
x(x + 16) - 12(x + 16) = 0
(x + 16)(x - 12) = 0
x + 16 = 0
x = -16
Or
x - 12 = 0
x = 12
Since, x being a number,
Therefore,
When x = -16 then
x + 4 = -16 + 4 = -12
And when x = 12 then
x + 4 = 12 + 4 = 16
Thus, two consecutive number be either 12, 16 or -16, -12.
APPEARS IN
संबंधित प्रश्न
Solve the following quadratic equations by factorization:
a2x2 - 3abx + 2b2 = 0
The sum of two number a and b is 15, and the sum of their reciprocals `1/a` and `1/b` is 3/10. Find the numbers a and b.
Three consecutive positive integers are such that the sum of the square of the first and the product of other two is 46, find the integers.
Solve:
(a + b)2x2 – (a + b)x – 6 = 0; a + b ≠ 0
Find two consecutive multiples of 3 whose product is 648.
A farmer wishes to grow a 100m2 rectangular vegetable garden. Since he was with him only 30m barbed wire, he fences 3 sides of the rectangular garden letting the compound of his house to act as the 4th side. Find the dimensions of his garden .
Solve the following quadratic equation by factorisation:
2x2 + ax - a2 = 0 where a ∈ R.
Solve (x2 + 3x)2 - (x2 + 3x) -6 = 0.
Solve the following by reducing them to quadratic form:
`sqrt(y + 1) + sqrt(2y - 5) = 3, y ∈ "R".`
Solve the equation:
`6(x^2 + (1)/x^2) -25 (x - 1/x) + 12 = 0`.