Math Problem Statement

Find three consecutive integers such that the product of the second and the third is 4 greater than ten times the first

Solution

Let the three consecutive integers be xx, x+1x+1, and x+2x+2.

According to the problem, the product of the second and third integers is 4 greater than 10 times the first integer. This gives us the equation:

(x+1)(x+2)=10x+4(x+1)(x+2) = 10x + 4

Step 1: Expand both sides

On the left-hand side, expand (x+1)(x+2)(x+1)(x+2):

(x+1)(x+2)=x2+3x+2(x+1)(x+2) = x^2 + 3x + 2

Now substitute this into the equation:

x2+3x+2=10x+4x^2 + 3x + 2 = 10x + 4

Step 2: Rearrange the equation

Move all terms to one side:

x2+3x+210x4=0x^2 + 3x + 2 - 10x - 4 = 0

Simplify the equation:

x27x2=0x^2 - 7x - 2 = 0

Step 3: Solve the quadratic equation

We solve x27x2=0x^2 - 7x - 2 = 0 using the quadratic formula. The quadratic formula is:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

For our equation x27x2=0x^2 - 7x - 2 = 0, the coefficients are:

  • a=1a = 1
  • b=7b = -7
  • c=2c = -2

Substitute these values into the quadratic formula:

x=(7)±(7)24(1)(2)2(1)x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(-2)}}{2(1)} x=7±49+82x = \frac{7 \pm \sqrt{49 + 8}}{2} x=7±572x = \frac{7 \pm \sqrt{57}}{2}

Since 57\sqrt{57} is not a perfect square, the exact solution involves irrational numbers, but we can approximate it:

x7±7.552x \approx \frac{7 \pm 7.55}{2}

This gives two approximate solutions: x17+7.552=7.275x_1 \approx \frac{7 + 7.55}{2} = 7.275 x277.552=0.275x_2 \approx \frac{7 - 7.55}{2} = -0.275

Since we are looking for integers, we round xx to the nearest integer. The integer closest to x1=7.275x_1 = 7.275 is x=7x = 7, and closest to x2=0.275x_2 = -0.275 is x=0x = 0.

Checking the solutions:

  • For x=7x = 7:

    • The integers are 7, 8, and 9.
    • The product of the second and third integers is 8×9=728 \times 9 = 72.
    • Ten times the first integer is 10×7=7010 \times 7 = 70, and adding 4 gives 70+4=7470 + 4 = 74.
    • This does not satisfy the condition.
  • For x=0x = 0:

    • The integers are 0, 1, and 2.
    • The product of the second and third integers is 1×2=21 \times 2 = 2.
    • Ten times the first integer is 10×0=010 \times 0 = 0, and adding 4 gives 0+4=40 + 4 = 4.
    • This does not satisfy the condition either.

Since neither solution satisfies the condition exactly, it seems there may be an issue with the rounding approximation.

Would you like me to explore further or try a different approach? Here are some related questions:

  1. How do you solve quadratic equations using different methods?
  2. Can you explain how to approximate irrational numbers?
  3. What are the steps to expand binomial expressions like (x+1)(x+2)(x+1)(x+2)?
  4. Why does the quadratic formula work in all cases?
  5. How can we check for integer solutions to equations like these?

Tip: Always double-check results when dealing with approximations or rounding.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Consecutive Integers

Formulas

Quadratic equation formula: ax^2 + bx + c = 0
Expansion of binomials: (x+1)(x+2)
Quadratic formula: x = (-b ± √(b² - 4ac)) / 2a

Theorems

Quadratic Formula

Suitable Grade Level

Grades 8-10