Question

The sum of the squares of two numbers is 233 and one of the number is 3 less than twice the other number. Find the numbers.

Answer 1

SOLUTION :

Let the number be  x and the other number be (2x - 3) .

A.T.Q

x² + (2x - 3)² = 233

x² + 4x² + 9 - 12x = 233

5x² - 12x - 224 = 0

5x² - 40x + 28x - 224 = 0

[By middle term splitting method]

5x(x - 8) + 28(x - 8) = 0

(5x + 28) (x - 8) = 0

(5x + 28)  = 0  or (x - 8) = 0

5x = - 28  or x = 8  

x = - 28/5  or x = 8  

x ≠ - 28/5  because square of a number cannot be negative it is always positive .

Therefore , x = 8  

First number (x) = 8  

Second number =  (2x - 3) = (2 × 8 - 3) = 16 - 3 = 13  

Hence, the required numbers be 8 & 13 .

HOPE THIS  ANSWER WILL HELP YOU…..

Answer 2

Answer:

8 , 13

-5.6 , -14.2

Step-by-step explanation:

X^2 + (2X-3)^2 = 233

X^2 + 4X^2 -12X +9 = 233

5X^2 -12X - 224 = 0

5X^2 -40X + 28X -224 = 0

5X(X-8)+28(X-8) = 0

(5X+28)(X-8)=0

X = 8 then 2X-3 = 13

X= -28/5 = -5.6 then 2X-3 = -14.2