Advertisements
Advertisements
प्रश्न
The sum of two numbers is 9 and the sum of their squares is 41. Taking one number as x, form ail equation in x and solve it to find the numbers.
उत्तर
Sum of two numbers = 9
Let first number = x
then second number = 9 – x
Now according to the condition,
(x)2 + (9 - x)2 = 41
⇒ x2 + 81 - 18x + x2 - 41 = 0
⇒ 2x2 - 18x + 40 = 0
⇒ x2 - 9x + 20 = 0 ...(Dividing by 2)
⇒ x2 - 4x - 5x + 20 = 0
⇒ (x - 4) -5(x - 4) = 0
⇒ (x - 4) (x - 5) = 0
Either x - 4 = 0,
then x = 4
or
x - 5 = 0,
then x = 5
(i) If x = 4, then first number = 4
and second number = 9 - 4 = 5
(ii) If x = 5, then first number = 5
ans second number = 9 - 5 = 4
Hence numbers are 4 and 5.
APPEARS IN
संबंधित प्रश्न
Solve for x :
`3/(x+1)+4/(x-1)=29/(4x-1);x!=1,-1,1/4`
Solve the following quadratic equations by factorization:
x2 - x - a(a + 1) = 0
Solve the following quadratic equations by factorization:
abx2 + (b2 – ac)x – bc = 0
Solve the following quadratic equations by factorization:
`(x-3)/(x+3 )+(x+3)/(x-3)=2 1/2`
The sum of the squares two consecutive multiples of 7 is 1225. Find the multiples.
Factorise : m2 + 5m + 6.
Two natural numbers differ by 4. If the sum of their square is 656, find the numbers.
There are three consecutive positive integers such that the sum of the square of the first and the product of other two is 154. What are the integers?
(x – 3) (x + 5) = 0 gives x equal to ______.
Using quadratic formula find the value of x.
p2x2 + (p2 – q2)x – q2 = 0