Question

The sum of a number and its square is $\frac{63}{4}$, find the numbers.

Answer 1

SOLUTION :  

Let the number  be x and square of the number = x².

A.T.Q

x² + x = 63/4

4(x² +x )  = 63

4x² + 4x = 63

4x² + 4x -  63 = 0

4x² +18x - 14x - 63 = 0

[By middle term splitting method]

2x(2x + 9) - 7(2x + 9) = 0

(2x - 7) (2x + 9) = 0

(2x - 7)  = 0  (2x + 9) = 0

2x = 7  or  2x = - 9

x = 7/2 or  x = - 9/2  

Hence, the required number is 7/2  & - 9/2.  

HOPE THIS  ANSWER WILL HELP YOU..

Answer 2

Solution :

Let the number = x ,

square the number = x²

According to the problem given ,

Sum = 63/4

=> x + x² = 63/4

=> 4( x² + x ) - 63 = 0

=> 4x² + 4x - 63 = 0

Splitting the middle term , we get

=> 4x² + 18x - 14x - 63 = 0

=> 2x( 2x + 9 ) - 7( 2x + 9 ) = 0

=> ( 2x + 9 )( 2x - 7 ) = 0

=> 2x + 9 = 0 or 2x - 7 = 0

=> 2x = -9 or 2x = 7

=> x = -9/2 or x = 7/2

Therefore ,

Required number x = -9/2 or 7/2

••••