Quadratic Word Problems: Consecutive Integers, Class 10

Quick answer: Let consecutive integers be $x$ and $x + 1$; consecutive odd or even integers be $x$ and $x + 2$. Form the equation, write it as $ax^2 + bx + c = 0$, factorise by splitting the middle term, and reject any root that breaks the conditions (for example a negative value when the numbers must be positive). The product-306 problem gives 17 and 18, the odd-squares-290 problem gives 11 and 13, and the two-squares problem gives sides of 18 m and 12 m.

Writing consecutive numbers algebraically

Rule: If the question asks for positive numbers or a length, reject any negative root and say why.

Problem 1: product of two consecutive positive integers is 306

Step 1: Let the integers be $x$ and $x + 1$.

Step 2: Form the equation:

$x(x + 1) = 306 \implies x^2 + x - 306 = 0$

Step 3: Split the middle term. We need two numbers with product $-306$ and sum $+1$. Since $306 = 18 \times 17$, they are $+18$ and $-17$:

$x^2 + 18x - 17x - 306 = 0$

$(x + 18)(x - 17) = 0$

Step 4: So $x = -18$ or $x = 17$. Reject $-18$ because the integers must be positive.

Answer: The integers are 17 and 18. Check: $17 \times 18 = 306$.

Problem 2: two consecutive odd positive integers whose squares add to 290

Step 1: Let the integers be $x$ and $x + 2$.

Step 2: Form the equation and expand:

$x^2 + (x + 2)^2 = 290$

$2x^2 + 4x + 4 = 290 \implies 2x^2 + 4x - 286 = 0$

Step 3: Divide by 2:

$x^2 + 2x - 143 = 0$

Step 4: We need two numbers with product $-143$ and sum $+2$. Since $143 = 11 \times 13$, they are $+13$ and $-11$:

$(x + 13)(x - 11) = 0$

Step 5: So $x = -13$ or $x = 11$. Reject $-13$ because the integers must be positive.

Answer: The integers are 11 and 13. Check: $121 + 169 = 290$.

Problem 3: two squares with total area 468 m²

The sum of the areas of two squares is 468 m², and their perimeters differ by 24 m. Find the sides.

Step 1: Let the sides be $x$ m (larger) and $y$ m (smaller).

Step 2: Perimeter condition: $4x - 4y = 24$, so $x = y + 6$.

Step 3: Area condition: $x^2 + y^2 = 468$. Substitute $x = y + 6$:

$(y + 6)^2 + y^2 = 468$

$2y^2 + 12y + 36 = 468 \implies y^2 + 6y - 216 = 0$

Step 4: We need two numbers with product $-216$ and sum $+6$: they are $+18$ and $-12$:

$(y + 18)(y - 12) = 0$

Step 5: Reject $y = -18$, since a length cannot be negative. So $y = 12$ and $x = 18$.

Answer: The sides are 18 m and 12 m. Check: $324 + 144 = 468$ and $4 \times 18 - 4 \times 12 = 24$.

Summary table

Exam tips

• State clearly what $x$ stands for before writing the equation.
• Write the reason for rejecting a root in words, for example "rejected, as the integers are positive".
• Give both numbers in the final answer, not just the value of $x$.
• Check your answer by substituting back into the original condition.

Common mistakes

• Writing consecutive odd numbers as $x$ and $x + 1$. Consecutive odd (or even) numbers differ by 2.
• Expanding $(x + 2)^2$ as $x^2 + 4$, forgetting the middle term $4x$.
• Forgetting to bring every term to one side before factorising.
• Keeping a negative root when the question asks for positive numbers or lengths.

Frequently Asked Questions

How do you write two consecutive odd numbers in algebra?

Write them as x and x + 2, where x is odd. They differ by 2, not 1.

Which two consecutive positive integers have a product of 306?

17 and 18, from x² + x − 306 = 0, which factorises as (x + 18)(x − 17) = 0.

Why do we reject the negative root?

Because the question asks for positive integers or lengths, and a negative value does not satisfy that condition.

What are two consecutive odd positive integers whose squares add up to 290?

11 and 13, since 121 + 169 = 290.

Can these problems be solved with the quadratic formula?

Yes. The quadratic formula gives the same roots; splitting the middle term is just quicker when the numbers factorise neatly.