Question

the difference between the squares of two consecutive integers is 47 find the numbers
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Answer 1

Answer:

let the first number be "x"

so the other no is "x+1"

let the square of fierst number be x^2

let the square of other number be (x+1)^2

ATQ the square of other no is 47-x^2

Thus x^2+2x+1=47-x^2

   2x^2+ 2x=46

2(x^2+x)=46

x^2+x=23

x^2+x-23=0

NOW Solving    x^2+x-23 = 0 by the Quadratic Formula .

According to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :

                                     

           - B  ±  √ B2-4AC

 x =   ————————

                     2A

 In our case,  A   =     1

                     B   =    -1

                     C   =  -23

Accordingly,  B2  -  4AC   =

                    1 - (-92) =

                    93

Applying the quadratic formula :

              1 ± √ 93

  x  =    —————

                   2

 √ 93   , rounded to 4 decimal digits, is   9.6437

So now we are looking at:

          x  =  ( 1 ±  9.644 ) / 2

Two solutions were found :

x =(1-√93)/2=-4.322

x =(1+√93)/2= 5.322

Answer 2

Answer:

answer is 23 and 24

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