Advertisements
Advertisements
प्रश्न
The sum of the square of 2 consecutive odd positive integers is 290.Find them.
Advertisements
उत्तर
Let the numbers be X and X+ 2. Then as per question,
X2 + (X+ 2)2= 290
⇒ 2x2 + 4X - 286 = 0
⇒ X2 + 2X - 143 = 0
⇒ X2 + 13X - 11X -143 = 0
⇒ X (X +13)-11(X +13) = 0
⇒ (X+13 )(X-11) = 0. X can't be negative number as its natural number.
Hence, X = 11and X+ 2= 13.
APPEARS IN
संबंधित प्रश्न
Solve the following quadratic equation by factorization method : `3x^2-29x+40=0`
Find the consecutive even integers whose squares have the sum 340.
A two digit number is 4 times the sum of its digits and twice the product of its digits. Find the number.
Find the value of k for which the following equations have real and equal roots:
\[x^2 - 2\left( k + 1 \right)x + k^2 = 0\]
Solve the following equation: 2x2 - x - 6 = 0
Solve the following equation: a2x2 - 3abx + 2b2 = 0
Solve the following equation: `1/("x" - 1) + 2/("x" - 1) = 6/"x" , (x ≠ 0)`
The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages was 124. Determine their present ages.
Solve equation using factorisation method:
(2x – 3)2 = 49
If the sum of the roots of the quadratic equation ky2 – 11y + (k – 23) = 0 is `13/21` more than the product of the roots, then find the value of k.
