Question

Two numbers are in the ratio 2 : 3 , if the sum of their squares is 117. Find the greater number.​

Answer 1

Answer:

Let the larger integer be a and the smaller integer be b . Then we are given:

a−ba2+b2=3=117(1)(2)

Squaring (1) , we have:

a2−2ab+b2=9⇒2ab=(a2+b2)−9(3)

Substituting (2) into (3) , we get

2ab=117−9⇒ab=54(4)

Then, from (1) we have b=a−3, which we can substitute into (4) to get:

a(a−3)=54(5)

From here, we can proceed a couple of ways. We can solve the quadratic

a2−3a−54=0

or we can take an educated guess. We first note from (5) that a and a−3 must bracket 54−−√. Since 54−−√ is between 7 and 8 , the only possible integer values for a are 8, 9, and 10. By trial and error (or noting that 54=6×9 ) we quickly find that a=9 solves (5) . Therefore our two numbers are 6 and 9.

Checking our result by substituting into (1) and (2) :

9−6=392+62=81+36=117✓✓

Answer 2

Answer:

Let the two numbers be = 2x , 3x

given sum of squares = 117

• ( 2x ) ² + ( 3x ) ² = 117
• 4x² + 9x² = 117
• 13x² = 117
• x² = 117/13
• x²= 9
• x = 3

Now the numbers = 2 * 3 = 6

3* 3 = 9

Greater number = 9