Quadratic Equations: Statement Questions (Part 1)

This lesson works through three statement (word) problems that lead to quadratic equations. For each one we name the unknown, build the equation, reduce it to standard form, and solve by factorisation, keeping only the root that makes sense.

Two consecutive odd positive integers

Find two consecutive odd positive integers the sum of whose squares is $290$.

Let the first odd integer be $x$, so the next odd integer is $x + 2$. The condition gives:

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

Expanding $(x+2)^2 = x^2 + 4x + 4$:

$2x^2 + 4x + 4 = 290$

$2x^2 + 4x - 286 = 0$

Divide both sides by $2$:

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

The product is $-143$ and the sum is $2$, which splits as $+13$ and $-11$:

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

So $x = -13$ or $x = 11$. Since $x$ must be a positive integer, $x = 11$. The next odd integer is $11 + 2 = 13$.

The numbers are $11$ and $13$. Check: $11^2 + 13^2 = 121 + 169 = 290$.

Shyam's marks in mathematics and English

The sum of Shyam's marks in mathematics and English is $30$. If he had got $2$ marks more in mathematics and $3$ marks less in English, the product of the marks would have been $210$. Find his marks.

Let his mark in mathematics be $x$. Then his mark in English is $30 - x$.

New marks. Two more in mathematics gives $x + 2$. Three less in English gives $(30 - x) - 3 = 27 - x$.

The product condition:

$(x + 2)(27 - x) = 210$

Expanding:

$27x - x^2 + 54 - 2x = 210$

$-x^2 + 25x + 54 = 210$

Multiply through by $-1$ and bring everything to one side:

$x^2 - 25x + 156 = 0$

The product is $156$ and the sum is $25$, splitting as $13$ and $12$:

$(x - 13)(x - 12) = 0$

So $x = 13$ or $x = 12$.

If $x = 13$

Mathematics $= 13$, English $= 30 - 13 = 17$.

If $x = 12$

Mathematics $= 12$, English $= 30 - 12 = 18$.

Both pairs are valid. Check $x = 13$: $(15)(14) = 210$.

Sides of a rectangular field

The diagonal of a rectangular field is $60$ m more than the shorter side. The longer side is $30$ m more than the shorter side. Find the sides.

Let the shorter side be $x$. Then the longer side is $x + 30$ and the diagonal is $x + 60$.

For the right triangle formed by the two sides and the diagonal, by Pythagoras:

$(x + 60)^2 = (x + 30)^2 + x^2$

Expanding each square:

$x^2 + 120x + 3600 = x^2 + 60x + 900 + x^2$

The $x^2$ on the left cancels with one $x^2$ on the right:

$120x + 3600 = x^2 + 60x + 900$

Bringing all terms to one side:

$x^2 - 60x - 2700 = 0$

The product is $-2700$ and the sum is $-60$, splitting as $-90$ and $+30$:

$(x - 90)(x + 30) = 0$

So $x = 90$ or $x = -30$. A length cannot be negative, so $x = 90$.

The shorter side (breadth) is $90$ m and the longer side (length) is $90 + 30 = 120$ m. Check: $90^2 + 120^2 = 8100 + 14400 = 22500 = 150^2$, and the diagonal $150 = 90 + 60$.

Key takeaways

• Name the unknown, then write each condition of the problem as an equation in that unknown.
• Always reduce to the standard form $ax^2 + bx + c = 0$ before factorising.
• Solve by splitting the middle term, then reject any root that breaks a real-world constraint (such as a negative length or a non-positive integer).