A math teacher gave three consecutive positive integers whose sum ofthe square o...

Questions

ericcaber1983 @ericcaber1983

September 2022 1 2 Report

A math teacher gave three consecutive positive integers whose sum of
the square of first and the product of the other two is 46. Find the three
positive integers.

Answers & Comments

Hajrudin

Answer:

The three consecutive positive integers are 4, 5 and 6

Step-by-step explanation:

Given:

First number = x

Second number = x+1

Third number = x+2

Equation:

$46 = x^{2} + (x + 1)(x + 2)$

Solution:

$46 = x {}^{2} + (x + 1)(x + 2) \\ 46 = x {}^{2} + x {}^{2} + 2x + x + 2 \\ 46 = 2x {}^{2} + 3x + 2 \\ 44 = 2x {}^{2} + 3x$

Move 44 to the other side as change the sign

$0 = 2x {}^{2} + 3x - 44$

Write 3x as a difference

$0 = 2x {}^{2} + 11x - 8x - 44$

Factor out x and -4

$0 = x(2x + 11) - 4(2x + 11) \\ (2x + 11)(x - 4) = 0$

$x - 4 = 0 \\ x = 4 \\$

$2x + 11 = 0 \\ 2x = - 11 \\ x = - \frac{11}{2}$

$x = 4 \: or \: - \frac{11}{2}$

Choose the positive answer

$x = 4$

Substitute

1st number = 4

2nd number = 5

3rd number = 6

Checking:

$46 = 4 ^{2} + (5)(6) \\ 46 = 16 + 30$

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Questions

Answers & Comments

The three consecutive positive integers are 4, 5 and 6

Given:

Equation:

Solution:

Checking:

Add an Answer

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