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

Let the three consecutive integers be \( x \), \( x+1 \), and \( x+2 \).

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:

\[
(x+1)(x+2) = 10x + 4
\]

### Step 1: Expand both sides
On the left-hand side, expand \( (x+1)(x+2) \):

\[
(x+1)(x+2) = x^2 + 3x + 2
\]

Now substitute this into the equation:

\[
x^2 + 3x + 2 = 10x + 4
\]

### Step 2: Rearrange the equation
Move all terms to one side:

\[
x^2 + 3x + 2 - 10x - 4 = 0
\]

Simplify the equation:

\[
x^2 - 7x - 2 = 0
\]

### Step 3: Solve the quadratic equation
We solve \( x^2 - 7x - 2 = 0 \) using the quadratic formula. The quadratic formula is:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For our equation \( x^2 - 7x - 2 = 0 \), the coefficients are:
- \( a = 1 \)
- \( b = -7 \)
- \( c = -2 \)

Substitute these values into the quadratic formula:

\[
x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(-2)}}{2(1)}
\]
\[
x = \frac{7 \pm \sqrt{49 + 8}}{2}
\]
\[
x = \frac{7 \pm \sqrt{57}}{2}
\]

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

\[
x \approx \frac{7 \pm 7.55}{2}
\]

This gives two approximate solutions:
\[
x_1 \approx \frac{7 + 7.55}{2} = 7.275
\]
\[
x_2 \approx \frac{7 - 7.55}{2} = -0.275
\]

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

### Checking the solutions:

- For \( x = 7 \):
  - The integers are 7, 8, and 9.
  - The product of the second and third integers is \( 8 \times 9 = 72 \).
  - Ten times the first integer is \( 10 \times 7 = 70 \), and adding 4 gives \( 70 + 4 = 74 \).
  - This does **not** satisfy the condition.

- For \( x = 0 \):
  - The integers are 0, 1, and 2.
  - The product of the second and third integers is \( 1 \times 2 = 2 \).
  - Ten times the first integer is \( 10 \times 0 = 0 \), and adding 4 gives \( 0 + 4 = 4 \).
  - 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 \((x+1)(x+2)\)?
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.

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

Solve a challenging algebra problem involving three consecutive integers. Learn how to use quadratic equations and the quadratic formula to find integers that satisfy the condition where the product of the second and third integers is 4 greater than ten times the first. Suitable for students in Grades 8-10.

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