problem?????

Question

one number is 5 greater than another number . if the sum of their squares is 5 times the square of the smaller number , what are the numbers?
if a number is subtracted from twice its reciprocal , the result is
-23/5. what is the number?
what are the techniques in answering this kind of problem?

Answer 1

The step shall inlove assuming some value x to the number and then proceding based on the steps.It shall be clearer with the examples

1) Let the smaller number be x
then larger number is x+5
sum of squares = x^2+(x+5)^2 = x^2 + x^2+10x+25 or 2x^2+10x+25
5 times the square of smaller number = 5x^2
both are same

5x^2= 2x^2+10x+25
or add -5x^2 on both sides

-3x^2+10x+25 =0
or 3x^2-10x-25 =0
or 3x^2-15x+5x-25=0
or 3x(x-5)+5(x-5) = 0
or (x-5)(3x+5) =0
x = 5 or -5/3

so the 2 numbers are 5 and 10
or -5/3 and 10/3
substitute the values and check for ans

I could have chosen larger number as x then I would have to compute 5(x-5)^2 making approach more complex.

2)
Let number be x
twice the reciprocal= 2/x
(2/x)-x = - 23/5 ( number subtracted from twice reciprocal = (2/x)-x

or
2-x^2 = -23/5 x
or 10 - 5x^2 = -23 x
ot 5x^2-23x -10 =0
or 5x^2 -25x+2x-10 = 0
or 5x(x-5)+2(x-5) = 0
or (x-5)(5x+2) = 0
x= 5 or -2/5

Answer 2

1. Let x =the smaller number
x+5= the bigger than number

Given: sum of their squares =5 times the square of the smaller number. Express this as an equation:

x^2+(x+5)^2=5x^2

Simplify:

x^2+x^2+10x+25=5x^2
x^2+x^2-5x^2+10x+25=0
-3x^2+10x+25=0

Multiply the equation by (-1):

3x^2-10x-25=0

Factor this equation:

(x -5)(3x+5)=0

Therefore,
(x-5)=0
x=5 (the smaller number)
x+5 =10(the bigger number)
5^2+10^2=5(5^2)
25+100=5(25)
125=125

and 3x+5=0
3x=-5
x=-5/3 (the smaller number)
x+5 =-5/3+5
=(-5+15)/3
=10/3

Check: (-5/3)^2+(10/3)^2=5(-5/3)^2
25/9+100/9=5(25)/9
25+100=5(25)
125=125.

2. Let x= the number

Its reciprocal is 1/x

Given: 2(1/x)-x=-23/5

2-x^2=(-23/5)x
-x^2+2=-23x/5

Multiply both sides by 5:

-5x^2+10=23x
-5x^2-23x+10=0

Multiply by (-1):

5x^2+23x-10=0

Answer 3

The number that is the "focus" of the problem can be taken as 'x' and other subsequent numbers as 'y', 'z', etc etc. Then you relate them with the equations given. For example I will solve the 2 probs u mentioned:

Let the greater no. - x
smaller no. - y
x = 5 + y ---------------------------eq# 1
(x)^2 +(y)^2 = 5(y)^2 -------------eq #2
we know, x= 5+y
Then eq# 2 becomes:
(5+y)^2 + (y)^2 = 5(y)^2
(25 + 10y + y^2) + y^2 = 5y^2
25 + 10y = 3y^2
or, 3y^2 + 10y + 25 = 0 ----------------- eq# 3
Now use quadratic equation to solve for y:
If you write eq# 3 as:
ay^2 +by + c = 0
then y = {-b +- sqrt(b^2 - 4ac)} / (2a)--------- QUADRATIC EQ.
So y for eq# 3 is :
y = {-10 +- sqrt(100 -300} / 6
y = 5/3 [1 +- sqrt2i]


For the second q:

number - x
reciprocal - 1/x
{2(1/x) -x} = -23/5
{(2-x^2)/x} = -23/5
10 - 5x^2 = -23x
-5x^2 +23x +10 = 0
Now use the quadratic equation as shown in the previous question to solve for x
x should be 75.2 and -70.6


P.N>>> I am not sure of the answers as I just did it in my head but am sure of the procedures.

Answer 4

one is 5 & other is 10 10 is 5 greater than 5 & sum of their squares is 125 is 5 times of squares 5

Answer 5

read every line with lot of concentration try to derive mathematical equation from each line. you will get no. of equation and now by method of substitution or elimination try to find out the variable
as in given question lets consider the no. as x
if one no is x then y (another no.) is x+5
so we get y= x+5 -----------(1)
now frm 2nd line we get another equation that is
x^2 + y^2 = 5x^2
now we can substitute the value of y in terms of x frm eq. 1 here
we get
x^2 + (x+5)^2 =5x^2
x^2 + x^2 +10x + 25 =5x^2
2x^2 - 5x^2 + 10x +25 =0
3x^2 - 10x- 25=0
3x^2 -15x +5x -25=0
3x(x-5) +5(x-5) =0
(3x+5) (x-5) = 0
x=5, -5/3
and coress. y =10 and 10/3


now second question
consider a no. x
equations comes as
2/x- x = -23/5
10- 5x^2= -23x
5x^2- 23x- 10= 0
5x^2 -25x +2 x- 10=0
5x(x-5) +2 (x-5) =0
(5x+2) (x-5)= 0
x=5, -2/5
hope this will help u out in answering these type of question
my advice: practice is the key method to learn any mathod in maths

Answer 6

let bgger number be x. then x=5+y. now x*x +y*y= 5y*y. substitute for x so (5+y)(5+y) +y*y=5y*y 25+10y+2Y*y=5y*y
25+10y-3y*y=0 25+15y-5y-3y*y=0 5(5+3y)-y(5+3y)=0 (5-y)(5+3y)=0 y=5 or y=-5/3 so now you can find x.
i think you can also solve the other in the same way. x-2/x=-23/5
then you get a quadratic equation which you must solve and get two values. the technique is to substitute the numbers with x and y and then solve.

Answer 7

Hope you know to write in figures or numbers.

Answer for the first one is as follows...

x^2+(x+5)^2=5x^2.(if you have doubt in this inform again)
Just expand the possible factors
you will get 2x^2+10x+25=5x^2

on transposing it, we get
3x^2-10x-25=0
3x^2-15x+5x-25=0
after solving it you get,x=5 or x=-5/3
If you are not satisfied then verify it by substituting the values.

Thus the basic technique is to expand the possible factors & solve the quadratic equation

Answer 8

The technique is to wirte the facts down as equasions and then solve them. For example, the first problem is:

x = y + 5
x^2 + y^2 = 5y^2

solve the equasions and you'll have the answer

Answer 9

1) 5 & 10
2) 5

write the equations

Answer 10

You must try to put the information in to a equations. Make your two Numbers "x" and "y"