Question

the difference of squares of two numbers is 88. if the larger number is 5 less than twice the smaller number, then find the two numbers.​

Answer 1

Step-by-step explanation:

Let two numbers be x and y and x(larger) y (smaller).

then, x^2-y^2 = 88 ----(2)

A/q ,

x = 2y - 5 ----(1)

Substitute value of x in given eqn 2,

(2y-5) ^2 -y^2 = 88

=>4y^2 + 25 - 20 y - y^2 = 88

=>3y^2 - 20y +25 -88 = 0

=> 3y^2 - 20y - 63 = 0

=>3y^2 - 27y +7y-63 =0

=>3y(y-9) +7(y-9) = 0

( 3y + 7)(y-9) = 0

=> y=9 ( right)

or, y = -7/3 (neglected)

Since Y=9,

Therefore,

x= 2*9 - 5 = 18 - 5 = 13

Hence,the required two numbers are 13 and 9.

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Answer 2

Method 1

let the smaller no.be x

then other no.=2x-5

now, according to question

(2x-5)^2-x^2=88

4x^2+25-20x-x^2=88

3x^2-20x+25=88

3x^2-20x-63=0

solving this equation u will get 13 and 9 as the answer.

method 2

let the first no. be x and second be y

x^2-y^2=88....(.2)

x=2y-5...,(1)

put the value of x

in 2nd

(2y-5)^2-y^2=88

4y^2+25-20y-y^2-88=0

3y^2 -20y-63=0

solving this u will get the same 13and 9 as the required number.

Verification

according to question ,the difference of difference of the numbers must be 88.

let us check this.

(13)^2-(9)^2=169-81=88,which is the requirement

hence the solution is exactly right.

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