Question

The difference of two positive integers is 3 and the sum of their squares is 89. Taking the

smaller integer as x form a quadratic equation and solve it to find the integers. The integers

are.

​

Answer 1

Answer:

since the small integer is x

other integer is x-3

as given, x²+(x-3)²=89

=> x²+x²+9-6x=89

=> 2x²-6x-80=0

=> x²-3x-40=0

=> x²-8x+5x-40=0

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

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

=>x=8(or)-5

but the integer is positive

so, x=8

so, other integer is 5

Answer 2

The two positive integers found after solving the quadratic equation formed from the given problem are 5 and 8.

Step-by-step Explanation

Given:

• The difference between two positive integers is 3.

• The sum of the squares of the two positive integers is 89.
• The smallest integer can be taken as x.

To be found:

To solve the quadratic equation formed from the given problem and find both the integers.

Solution:

Since it is given, take the smallest integer as x.

We know that the difference between two positive integers is 3, so the second integer will be equal to x-3.

Now, using the second condition, it is given that the sum of the squares of the two positive integers is 89.

$\implies x^{2} +(x-3)^{2} =89$

Rearranging, we get

$\implies x^{2} +(x-3)^{2} -89=0\\\implies x^{2} + x^{2} +9-6x-89=0$         $[\because (a-b)^{2} =a^{2} +b^{2} -2ab]$

Simplifying, we get

$2x^{2} -6x-80=0$

Factorising, we get

$2x^{2} -16x+10x-80=0\\\implies 2x(x-8)+10(x-8)=0\\\implies (x-8)(2x+10)=0$

Solving, we get

$x-8=0$ and $2x+10=0$

$\implies x=8$ and $x=\frac{-10}{2}=-5$

It was already given that the integers are positive, so $x=8$

Also, $x-3=8-3=5$

Hence, the two positive integers are 5 and 8 respectively.

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