Advertisements
Advertisements
प्रश्न
The difference of squares of two number is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers.
उत्तर
Let the smaller numbers be x
Then according to question,
The larger number be = 2x - 5, then
(2x - 5)2 - x2 = 88
4x2 - 20x + 25 - x2 - 88 = 0
3x2 - 20x - 63 = 0
3x2 - 27x + 7x - 63 = 0
3x(x - 9) + 7(x - 9) = 0
(x - 9)(3x + 7) = 0
x - 9 = 0
x = 9
Or
3x + 7 = 0
3x = -7
x = -7/3
Since, x being a positive integer so, x cannot be negative,
Therefore,
When x = 9 then larger number be
2x - 5 = 2(9) - 5 = 18 - 5 = 13
Thus, two consecutive number be either 9, 13.
APPEARS IN
संबंधित प्रश्न
Solve the following quadratic equation by factorization method: 9x2-25 = 0
The sum of the squares to two consecutive positive odd numbers is 514. Find the numbers.
Solve the following quadratic equations by factorization: \[\frac{5 + x}{5 - x} - \frac{5 - x}{5 + x} = 3\frac{3}{4}; x \neq 5, - 5\]
Solve the following quadratic equations by factorization:
\[3\left( \frac{3x - 1}{2x + 3} \right) - 2\left( \frac{2x + 3}{3x - 1} \right) = 5; x \neq \frac{1}{3}, - \frac{3}{2}\]
Write the number of real roots of the equation x2 + 3 |x| + 2 = 0.
If one of the equation x2 + ax + 3 = 0 is 1, then its other root is
Solve the following by reducing them to quadratic form:
`sqrt(y + 1) + sqrt(2y - 5) = 3, y ∈ "R".`
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.
A person was given Rs. 3000 for a tour. If he extends his tour programme by 5 days, he must cut down his daily expenses by Rs. 20. Find the number of days of his tour programme.
Solve the following equation by factorisation :
`sqrt(3)x^2 + 10x + 7sqrt(3)` = 0
