Question

The sum of the square of two number is 34.The first number is one less than twice the second number.Find the number

Answer 1

Step-by-step explanation:

Given:-

• The sum of the square of two numbers is 34.

• The first number is one less than twice the second number.

To Find:-

• The two numbers.

Solution:-

Case 1:-

Let the second number be x

As, the first number is one less than twice the second number

The first number = 2x - 1

Case 2:-

The sum of squares of the two numbers = 34.

$\implies \sf x^2 + (2x - 1)^2 = 34$

$\implies \sf x^2 + {(2x)}^{2} + {(1)}^{2} - 2(2x)(1) = 34$

$\implies \sf x^2 + {4x}^{2} + 1 - 4x = 34$

$\implies \sf 5x^2 - 4x +1 - 34 = 0$

$\implies \sf 5x^2 - 4x - 33 = 0$

$\implies \sf 5x^2 - 15x + 11x - 33 = 0$

$\implies \sf 5x(x - 3) + 11(x - 3) = 0$

$\implies \sf (5x + 11) (x - 3)= 0$

$\implies \sf 5x + 11 = 0 \:\:\:( or) x - 3 = 0$

$\implies \sf x = \dfrac{-11}{3} , 3$

Since, the two numbers are natural.

$\sf \implies x = 3$

The second number = x = 3

The first number = 2x - 1 = 2(3) - 1 = 5

$\underline{\boxed{\sf \therefore The \: two \: numbers \: are \: 3 \: and \: 5}}$

Answer 2

Step-by-step explanation:

Let second no. = x

then first no. = 2x - 1

sum of two square no. = 34

(2x - 1)^2 + x^2 = 34

4x^2 - 4x + 1 + x^2 = 34

5x^2 - 4x + 1 = 34

5×^2 - 4x + 1 - 34 = 0

5x^2 - 4x - 33 = 0

5x^2 - 15x + 11x - 33 = 0

5x ( x - 3 ) + 11 ( x - 3) = 0

( 5x + 11 ) ( x - 3 ) = 0

5x + 11 = 0 x - 3 = 0

5x = - 11 x = 3

x = - 11 / 5

second no. = 3

First no. = 2x - 1 = 2 × 3 - 1

= 6 - 1 = 5