Question

The sum of the squares of two positive numbers is 233 and one number is 3 less than twice the other number. Find the numbers. ​

Answer 1

Answer :-

$\: \: \boxed{\boxed{\rm{\mapsto \: \: \: Firstly \: let's \: understand \: the \: concept \: used}}}$

Here the concept of Quadratic Equations has been used. We see, that here two unknown values are given with a single given where one value depend on other. So, now using the standard form of Quadratic Equation, let's solve the question.

___________________________________________

★ Question :-

The sum of the squares of two positive numbers is 233 and one number is 3 less than twice the other number. Find the numbers.

_____________________________________________

★ Solution :-

Given,

» Sum of squares of two positive numbers = 233

» One number + 3 = 2 × Other Number

• Let one number be 'x'.

• Let another number be 'y'.

Then, according to the question :-

~ Case I :-

⌬ x + 3 = 2y

⌬ x = 2y - 3 ... (i)

~ Case II :-

⌬ x² + y² = 233 ... (ii)

From equation, (i) and (ii), we get,

⌬ (2y - 3)² + y² = 233

⌬ 4y² - 9 - 12y + y² = 233

⌬ 5y² - 12y + 9 = 233

⌬ 5y² - 12y + 9 - 233 = 0

⌬ 5y² - 12y - 224 = 0

⌬ 5y² - 40y + 28y - 224 = 0

⌬ 5y(y - 8) + 28(y - 8) = 0

⌬ (5y + 28)(y - 8) = 0

Here either (5y + 28) = 0 , or (y - 8) = 0

⌬ 5y + 28 = 0 or y - 8 = 0

⌬ 5y = -28 or y = 8

$\: \: </strong><strong>\</strong><strong>h</strong><strong>u</strong><strong>g</strong><strong>e</strong><strong>{</strong><strong>\longrightarrow \: \: \bf{y \: = \: </strong><strong>\</strong><strong>dfrac{-28}{5} \: \: or \: \: y \: = \: 8}</strong><strong>}</strong><strong>$

We know that, in question its clearly given that two numbers are positive.

But here (-28) / 5 is a negative number. So we are avoiding this.

$\: \: \huge{\sf{:\Longrightarrow \: \: y \: = \: 8}}$

• Hence, the other number = y = 8

Now using case (i), we get,

⌬ x = 2y - 3

⌬ x = 2(8) - 3

⌬ x = 16 - 3

$\: \: \huge{\sf{:\Longrightarrow \: \: x \: = \: 13}}$

• Hence, one number is = x = 13

$\: \: \boxed{\boxed{\rm{\mapsto \: \: Thus \: one \: number \: is \: \underline{13} \: and \: other \: number \: is \: \underline{8}}}}$

________________________________________

$\: \: \: \: \underline{\underline{\rm{\Longrightarrow \: \: \: Confused? \: Don't \: worry \: let's \: verify \: it}}}$

For verification, we need to simply apply the values we got into the equations we formed. Then,

~ Case I :-

=> x = 2y - 3

=> 13 = 2(8) - 3

=> 13 = 13

Clearly, LHS = RHS

~ Case II :-

=> x² + y² = 233

=> (13)² + (8)² = 169 + 64 = 233

=> 233 = 233

Clearly, LHS = RHS

Here both the conditions satisfy, so our answer is correct.

Hence, Verified.

__________________________________

$\: \: \boxed{\underline{\tt{\mapsto \: \: \: Let's \: understand \: more}}}$

• Polynomials are the equations formed using constant and variable terms which can be of any degree.

• Different forms of Polynomials are :-

• Linear Polynomial
• Quadratic Polynomial
• Cubic Polynomial
• Bi - Quadratic Polynomial

• Linear Equations are the equations formed using constant and variable terms but of single degree.

• Types of Linear Equations are :-

• Linear Equation in One Variable
• Linear Equation in Two Variable
• Linear Equation in Three Variable

Answer 2

Given :

• Sum of squares of 2 positive numbers = 233
• One number is 3 less than twice the other number

To find :

• Two numbers

Solution :

Let the first number be x

∴ Second number = 2x - 3

Now atq,

⇒ x² + (2x - 3)² = 233

⇒ x² + 4x² - 2 * 2x * 3 + 9 = 233

⇒ 5x² - 12x + 9 = 233

⇒ 5x² - 12x + 9 - 233 = 0

⇒ 5x² - 12x - 224 = 0

⇒ 5x²  - 40x + 28x - 224 = 0

⇒ 5x(x - 8) + 28(x - 8) = 0

⇒ (5x + 28) (x - 8) = 0

⇒ 5x + 28 = 0   or,  x - 8 = 0

⇒ 5x = - 28    or,   x = 8

∵ Both are positive numbers

∴ x = 8

∴ One numbers, x = 8

∴ Other number, 2x - 3 = 2(8) - 3 = 16 - 3 = 13

Therefore,

The numbers are 8 and 13