Question

The difference and the product of two numbers are 32 and 2145 respectively. their sum is:

Answer 1

Let the numbers be x and Y respectively
X-Y=32----(I)
XY=2145-----(ii)
Put X=2145/Y in eq (I)
2145-Y^2-32Y=0
Now solve for X and Y ,then add them

Answer 2

Answer:

Required sum is 98.

Step-by-step explanation:

Given difference and the product of two numbers are 32 and 2145.

Let first number be x and second number be y .

According to question,

$x - y = 32$ and $xy = 2145$

We know,

${(x + y)}^{2} \\ = {(x - y)}^{2} + 4xy \\ = {32}^{2} + 4 \times 2145 \\ = 1 024 + 8580 \\ = 9604$

So,

${(x + y)}^{2} = 9604 \\ x + y = \sqrt{9604} \\ x + y = 98$

Therefore sum of two numbers is 98.

This is a problem of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Know more about Algebra,

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