Question

Good Morning

Addition of two numbers is 23 while multiplication is 120. Find the addition of square of both the numbers.​

Answer 1

Let:-

The First number be x

Then:-

Second Number will be (23 - x)

Now according to question-

➨ x(23 - x ) = 120

➨ x² - 23x = 120

➨ x² - 23x + 120 = 0

➨ x² - 15x - 8x + 120 = 0

➨ x (x - 15) - 8(x - 15) = 0

➨ ( x - 8) (x - 15) = 0

➨ x = 8 and x = 15

Therefore:-

First number is 8 and second number is 15.

Now according to Question:-

➨ (8)² + (15)²

➨ 64 + 225

➨ 289

Hence:-

Addition of square of both numbers is 289.

Answer 2

Step-by-step explanation:

$\bf{\underline{\underline\red{Given:-}}}$

• The addition of two numbers is 23.

• The multiplication of numbers is 120

$\bf{\underline{\underline\blue{To\:Find:-}}}$

• The addition of square of the both numbers.

$\bf{\underline{\underline\green{Solution:-}}}$

The sum of the both numbers is 23

Let one of the numbers be x

The other number = 23 - x

As the multiplication of the numbers is 120.

$\rm \implies (x)(23 - x) = 120$

$\rm \implies23x - {x}^{2} = 120$

$\rm \implies23x - {x}^{2} - 120 = 0$

$\rm \implies - (23x - {x}^{2} - 120 )= 0$

$\rm \implies - 23x + {x}^{2} + 120 = 0$

$\rm \implies {x}^{2} - 23x + 120 = 0$

$\rm \implies {x}^{2} - 15x - 8x + 120 = 0$

$\rm \implies {x}(x - 15) - 8(x - 15)= 0$

$\rm \implies (x - 15) (x - 8)= 0$

$\rm \implies x= 15 \: , \: 8$

Therefore, The numbers are 15 and 8

We are asked to find the sum of squares of the numbers

$\rm \implies {15}^{2} + {8}^{2}$

$\rm \implies 225 + 64$

$\rm \implies 289$

The sum of squares of the numbers is 289.

$\bf{\underline{\underline\pink{Alternative\:Method:-}}}$

Let the numbers be x and y

The sum of numbers is 23

→ x + y = 23

The multiplication of numbers is 120

→ xy = 120

Sum of squares of numbers = x² + y²

According to the Identity

$\rm \implies(x + y) ^{2} = {x}^{2} + {y}^{2} + 2xy$

$\rm \implies(2 3) ^{2} = {x}^{2} + {y}^{2} + 2(120)$

$\rm \implies529 = {x}^{2} + {y}^{2} +240$

$\rm \implies529 - 240= {x}^{2} + {y}^{2}$

$\rm \implies {x}^{2} + {y}^{2} = 289$

$\underline{\boxed {\purple{ \rm \therefore Addition \: of \: squares \: of \: both \: numbers = 289}}}$