Question

The sum of the squares of three positive numbers that are consecutive multiples of 5 is 725. Find the three numbers.

yeah
myself hasini
10th class
india​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that,

• First multiple of 5 be x

• Second multiple of 5 be x + 5

• Third multiple of 5 be x + 10

According to statement, sum of the squares of three positive numbers that are consecutive multiples of 5 is 725.

So,

$\rm \: {x}^{2} + {(x + 5)}^{2} + {(x + 10)}^{2} = 725$

$\rm \: {x}^{2} + {x}^{2} + 25 + 10x + {x}^{2} + 100 + 20x = 725$

$\rm \: 3{x}^{2} + 30x + 125 = 725$

$\rm \: 3{x}^{2} + 30x + 125 - 725 = 0$

$\rm \: 3{x}^{2} + 30x - 600 = 0$

$\rm \: 3({x}^{2} + 10x - 200) = 0$

$\rm \: {x}^{2} + 10x - 200 = 0$

$\rm \: {x}^{2} + 20x - 10x - 200 = 0$

$\rm \: x(x + 20) - 10(x + 20) = 0$

$\rm \: (x + 20)(x - 10) = 0$

$\rm\implies \:x = 10 \: \: or \: \: x \: = \: - \: 20$

Hence,

• First multiple of 5 = 10

• Second multiple of 5 = 10 + 5 = 15

• Third multiple of 5 = 10 + 10 = 20

Verification :-

First multiple of 5 = 10

Second multiple of 5 = 15

Third multiple of 5 = 20

So, Consider

$\rm \: {10}^{2} + {15}^{2} + {20}^{2}$

$\rm \: = \: 100 + 225 + 400$

$\rm \: = \: 725$

So, sum of the squares of three positive numbers that are consecutive multiples of 5 is 725.

Hence, Verified

$\rule{190pt}{2pt}$

Additional information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac

Answer 2

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: find : \\ three \: consecutive \: multiples \: of \: 5 \\ \\ let \: the \: three \: consecutive \: \\ multiples \: of \: 5 \: be \\ 5x,5(x + 1),5(x + 2) \\ \\ according \: to \: given \: condition \\ (5x) {}^{2} + [5(x + 1)] {}^{2} + [5(x + 2)] {}^{2} = 725 \\ \\ 25x {}^{2} + 25(x + 1) {}^{2} + 25(x + 2) {}^{2} = 725 \\ \\ 25[x {}^{2} + (x + 1) {}^{2} + (x + 2) {}^{2} ] = 725 \\ \\ x {}^{2} + (x + 1) {}^{2} + (x + 2) {}^{2} = 29 \\ \\ x {}^{2} + x {}^{2} + 2x + 1 + x {}^{2} + 4x + 4 = 29 \\ \\ 3x {}^{2} + 6x - 24 = 0 \\ \\ x {}^{2} + 2x - 8 = 0 \\ \\ x {}^{2} + 4x - 2x - 8 = 0 \\ \\ x (x + 4) - 2(x +4 ) = 0 \\ \\ (x + 4)(x - 2) = 0 \\ \\ x = - 4 \: \: or \: \: x = 2 \\ \\ but \: as \: given \: numbers \: are \: positive \\ x = - 4 \: is \: absurd \\ \\ x = 2$

$so \: then \\ 5x = (5 \times 2) = 10 \\ \\ 5(x + 1) = 5(2 + 1) = 5 \times 3 = 15 \\ \\ 5(x + 2) = 5 \times (2 + 2) = 5 \times 4 = 20 \\ \\ thus \: required \: numbers \: are \\ 10,15 \: and \: 20$