Solved Examples

How to Use This Section

This is your practice powerhouse for Quadratic Equations. Below are 30+ fully worked problems spanning the whole chapter — factorisation, completing the square, the quadratic formula, the discriminant, and word problems — arranged roughly easy to hard.

How to study: Try each problem with the solution covered, then check the steps. Board marks are awarded step by step — show full working and state word-problem answers in words.

Keep these handy:

• Quadratic formula: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Discriminant $D = b^2 - 4ac$: $>0$ distinct real, $=0$ equal, $<0$ no real roots.
• Completing the square: add $(b/2)^2$.
• Always reject inadmissible roots in word problems.

Solved Examples

Example 1: Factorise

Solve $x^2 - 9x + 20 = 0$.

Solution:

1. Numbers with product 20, sum $-9$: $-4, -5$.
2. $(x - 4)(x - 5) = 0 \Rightarrow x = 4, 5$.

Final Answer: $x = 4, 5$.

Takeaway: Standard split-the-middle-term factorisation.

Example 2: Factorise ($a \neq 1$)

Solve $3x^2 + 10x + 3 = 0$.

Solution:

1. $a \times c = 9$; numbers 9, sum 10: 9 and 1.
2. $3x^2 + 9x + x + 3 = 3x(x+3) + 1(x+3) = (3x+1)(x+3) = 0$.
3. $x = -\dfrac{1}{3}, -3$.

Final Answer: $x = -\dfrac{1}{3}, -3$.

Takeaway: Use $a \times c$ for the product.

Example 3: Quadratic formula

Solve $x^2 - 3x - 10 = 0$ using the formula.

Solution:

1. $D = 9 + 40 = 49$.
2. $x = \dfrac{3 \pm 7}{2} = 5$ or $-2$.

Final Answer: $x = 5, -2$.

Takeaway: Formula confirms factorisation $(x-5)(x+2)$.

Example 4: Completing the square

Solve $x^2 + 8x + 9 = 0$.

Solution:

1. $x^2 + 8x = -9$; add $16$: $(x+4)^2 = 7$.
2. $x = -4 \pm \sqrt7$.

Final Answer: $x = -4 \pm \sqrt7$.

Takeaway: Add $(8/2)^2 = 16$ to both sides.

Example 5: Discriminant / nature

Find the nature of the roots of $3x^2 - 2x + 1 = 0$.

Solution:

1. $D = 4 - 12 = -8 < 0$.

Final Answer: No real roots.

Takeaway: $D < 0$ ⇒ no real roots.

Example 6: Equal roots — find $k$

Find $k$ so that $4x^2 + kx + 9 = 0$ has equal roots.

Solution:

1. $D = k^2 - 4(4)(9) = k^2 - 144 = 0$.
2. $k = \pm 12$.

Final Answer: $k = \pm 12$.

Takeaway: Equal roots ⇒ $D = 0$.

Example 7: Surd roots by formula

Solve $2x^2 - 7x + 4 = 0$.

Solution:

1. $D = 49 - 32 = 17$.
2. $x = \dfrac{7 \pm \sqrt{17}}{4}$.

Final Answer: $x = \dfrac{7 \pm \sqrt{17}}{4}$.

Takeaway: Non-square $D$ ⇒ surd roots; leave in exact form.

Example 8: Difference of squares

Solve $25x^2 - 9 = 0$.

Solution:

1. $(5x - 3)(5x + 3) = 0$.
2. $x = \pm\dfrac{3}{5}$.

Final Answer: $x = \pm\dfrac{3}{5}$.

Takeaway: $a^2 - b^2 = (a-b)(a+b)$.

Example 9: Clearing fractions

Solve $\dfrac{x}{x+1} + \dfrac{x+1}{x} = \dfrac{5}{2}$ (assume $x \neq 0, -1$).

Solution:

1. Combine: $\dfrac{x^2 + (x+1)^2}{x(x+1)} = \dfrac{5}{2} \Rightarrow 2(2x^2 + 2x + 1) = 5(x^2 + x)$.
2. $4x^2 + 4x + 2 = 5x^2 + 5x \Rightarrow x^2 + x - 2 = 0$.
3. $(x + 2)(x - 1) = 0 \Rightarrow x = -2, 1$.

Final Answer: $x = -2, 1$.

Takeaway: Clear denominators to reach a standard quadratic.

Example 10: Word problem — numbers

The difference of two numbers is 3 and the sum of their squares is 117. Find them.

Solution:

1. Let them be $x$ and $x - 3$: $x^2 + (x-3)^2 = 117 \Rightarrow 2x^2 - 6x - 108 = 0 \Rightarrow x^2 - 3x - 54 = 0$.
2. $(x - 9)(x + 6) = 0 \Rightarrow x = 9$ or $x = -6$.
3. Numbers: 9 and 6 (or $-6$ and $-9$).

Final Answer: 9 and 6 (or $-6$ and $-9$).

Takeaway: Both sign-pairs can be valid if positivity isn't required.

Example 11: Completing the square with $a \neq 1$

Solve $2x^2 + 5x - 3 = 0$ by completing the square.

Solution:

1. Divide by 2: $x^2 + \dfrac{5}{2}x - \dfrac{3}{2} = 0 \Rightarrow x^2 + \dfrac{5}{2}x = \dfrac{3}{2}$.
2. Half of $\dfrac{5}{2}$ is $\dfrac{5}{4}$, square $\dfrac{25}{16}$: $\left(x + \dfrac{5}{4}\right)^2 = \dfrac{3}{2} + \dfrac{25}{16} = \dfrac{49}{16}$.
3. $x + \dfrac{5}{4} = \pm\dfrac{7}{4} \Rightarrow x = \dfrac{1}{2}$ or $-3$.

Final Answer: $x = \dfrac{1}{2}, -3$.

Takeaway: Divide by $a$ first; keep fractions exact.

Example 12: Roots given, find equation

Form the quadratic equation whose roots are $3$ and $-\dfrac{1}{2}$.

Solution:

1. Sum $= 3 - \dfrac{1}{2} = \dfrac{5}{2}$; product $= 3 \times -\dfrac{1}{2} = -\dfrac{3}{2}$.
2. $x^2 - (\text{sum})x + (\text{product}) = x^2 - \dfrac{5}{2}x - \dfrac{3}{2} = 0$.
3. Multiply by 2: $2x^2 - 5x - 3 = 0$.

Final Answer: $2x^2 - 5x - 3 = 0$.

Takeaway: Use $x^2 - (\text{sum})x + (\text{product})$, then clear fractions.

Example 13: Train word problem

A train travels 480 km. If its speed were 8 km/h less, it would take 3 hours more. Find the speed.

Solution:

1. $\dfrac{480}{x - 8} - \dfrac{480}{x} = 3$.
2. $480x - 480(x - 8) = 3x(x - 8) \Rightarrow 3840 = 3x^2 - 24x \Rightarrow x^2 - 8x - 1280 = 0$.
3. $(x - 40)(x + 32) = 0 \Rightarrow x = 40$.

Final Answer: Speed $= 40$ km/h.

Takeaway: 'Less speed, more time' gives a similar time-difference equation.

Example 14: Distinct roots condition

For what $p$ does $x^2 - px + 16 = 0$ have distinct real roots?

Solution:

1. $D > 0$: $p^2 - 64 > 0$.
2. $p > 8$ or $p < -8$.

Final Answer: $p > 8$ or $p < -8$.

Takeaway: Distinct real roots ⇒ $D > 0$.

Example 15: Reciprocal equation

Solve $x - \dfrac{1}{x} = \dfrac{3}{2}$ (assume $x \neq 0$).

Solution:

1. Multiply by $x$: $x^2 - 1 = \dfrac{3}{2}x \Rightarrow 2x^2 - 3x - 2 = 0$.
2. $(2x + 1)(x - 2) = 0 \Rightarrow x = -\dfrac{1}{2}, 2$.

Final Answer: $x = -\dfrac{1}{2}, 2$.

Takeaway: Clear the denominator, then factorise.

Example 16: Garden path (area)

A rectangular garden 16 m by 10 m has a uniform path of width $x$ around it inside, leaving an inner area of 120 m². Find $x$.

Solution:

1. Inner dimensions: $(16 - 2x)$ and $(10 - 2x)$. Area: $(16 - 2x)(10 - 2x) = 120$.
2. $160 - 52x + 4x^2 = 120 \Rightarrow 4x^2 - 52x + 40 = 0 \Rightarrow x^2 - 13x + 10 = 0$.
3. $x = \dfrac{13 \pm \sqrt{169 - 40}}{2} = \dfrac{13 \pm \sqrt{129}}{2}$. Taking the smaller root, $x \approx 0.82$ m (the larger is inadmissible).

Final Answer: $x = \dfrac{13 - \sqrt{129}}{2} \approx 0.82$ m.

Takeaway: Subtract $2x$ from each dimension for an inside border.

Example 17: Sum and product of roots

If the roots of $x^2 - 6x + k = 0$ differ by 4, find $k$.

Solution:

1. Let roots be $\alpha, \beta$: $\alpha + \beta = 6$, $\alpha - \beta = 4$. So $\alpha = 5, \beta = 1$.
2. $k = \alpha\beta = 5 \times 1 = 5$.

Final Answer: $k = 5$.

Takeaway: Use sum and difference to find each root, then the product.

Example 18: Factorise with surds

Solve $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$.

Solution:

1. $a \times c = 10$; split 7 as $2 + 5$.
2. $\sqrt2 x^2 + 2x + 5x + 5\sqrt2 = \sqrt2 x(x + \sqrt2) + 5(x + \sqrt2) = (x + \sqrt2)(\sqrt2 x + 5) = 0$.
3. $x = -\sqrt2$ or $x = -\dfrac{5}{\sqrt2} = -\dfrac{5\sqrt2}{2}$.

Final Answer: $x = -\sqrt2, -\dfrac{5\sqrt2}{2}$.

Takeaway: Same factorisation method with surds; rationalise the answer.

Example 19: Two-digit number

A two-digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange. Find the number.

Solution:

1. Tens $= x$, units $= y$: $xy = 18$ and $(10x + y) - 63 = 10y + x \Rightarrow 9x - 9y = 63 \Rightarrow x - y = 7$.
2. From $x = y + 7$ and $xy = 18$: $(y+7)y = 18 \Rightarrow y^2 + 7y - 18 = 0 = (y+9)(y-2)$; $y = 2$, $x = 9$.
3. Number $= 92$.

Final Answer: 92.

Takeaway: Combine a digit-product and a digit-difference condition into one quadratic.

Example 20: Equal roots — find $k$ (a-coefficient)

Find $k$ so that $kx^2 - 2\sqrt5 x + 4 = 0$ has equal roots.

Solution:

1. $D = (2\sqrt5)^2 - 4(k)(4) = 20 - 16k = 0$.
2. $k = \dfrac{20}{16} = \dfrac{5}{4}$.

Final Answer: $k = \dfrac{5}{4}$.

Takeaway: $(2\sqrt5)^2 = 20$; set $D = 0$ and solve for $k$.

Example 21: Age problem

The sum of the ages of a mother and daughter is 45 years. Five years ago, the product of their ages was 124. Find their present ages.

Solution:

1. Let daughter $= x$, mother $= 45 - x$. Five years ago: $(x - 5)(40 - x) = 124$.
2. $40x - x^2 - 200 + 5x = 124 \Rightarrow -x^2 + 45x - 324 = 0 \Rightarrow x^2 - 45x + 324 = 0$.
3. $(x - 9)(x - 36) = 0 \Rightarrow x = 9$ (daughter) since mother $>$ daughter.

Final Answer: Daughter 9 years, mother 36 years.

Takeaway: Express one age via the sum; subtract 5 from both for the past.

Example 22: No real roots — show

Show that $x^2 + x + 1 = 0$ has no real roots.

Solution:

1. $D = 1 - 4 = -3 < 0$.
2. Negative discriminant ⇒ no real roots.

Final Answer: No real roots ($D = -3$).

Takeaway: A quick $D$ computation settles existence of real roots.

Example 23: Formula with large numbers

Solve $x^2 - 45x + 324 = 0$.

Solution:

1. $D = 2025 - 1296 = 729 = 27^2$.
2. $x = \dfrac{45 \pm 27}{2} = 36$ or $9$.

Final Answer: $x = 9, 36$.

Takeaway: Recognising $729 = 27^2$ keeps the roots rational.

Example 24: Pythagoras word problem

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution:

1. Let base $= x$, altitude $= x - 7$: $x^2 + (x - 7)^2 = 169$.
2. $2x^2 - 14x + 49 = 169 \Rightarrow x^2 - 7x - 60 = 0 = (x - 12)(x + 5)$; $x = 12$.
3. Base 12 cm, altitude 5 cm.

Final Answer: Base 12 cm, altitude 5 cm.

Takeaway: Use $a^2 + b^2 = c^2$ with the side relation.

Example 25: Roots ratio condition

If one root of $x^2 - 6x + k = 0$ is twice the other, find $k$.

Solution:

1. Let roots be $\alpha$ and $2\alpha$. Sum: $3\alpha = 6 \Rightarrow \alpha = 2$.
2. Product: $\alpha \cdot 2\alpha = 2\alpha^2 = k \Rightarrow k = 2(4) = 8$.

Final Answer: $k = 8$.

Takeaway: Use the ratio to parametrise roots, then sum and product.

Example 26: Reducible to quadratic

Solve $x^4 - 13x^2 + 36 = 0$.

Solution:

1. Let $y = x^2$: $y^2 - 13y + 36 = 0 = (y - 4)(y - 9)$.
2. $y = 4$ or $9$, so $x^2 = 4$ or $9$.
3. $x = \pm 2, \pm 3$.

Final Answer: $x = \pm 2, \pm 3$.

Takeaway: A 'biquadratic' becomes quadratic via $y = x^2$.

Example 27: Word problem — playground

The area of a right-triangular plot is 600 m². The base exceeds the height by 10 m. Find the base and height.

Solution:

1. Area $= \dfrac{1}{2} \times \text{base} \times \text{height}$. Let height $= x$, base $= x + 10$: $\dfrac{1}{2}x(x + 10) = 600$.
2. $x^2 + 10x - 1200 = 0 = (x + 40)(x - 30)$; $x = 30$.
3. Height 30 m, base 40 m.

Final Answer: Height 30 m, base 40 m.

Takeaway: Triangle area uses the $\dfrac{1}{2}$ factor; reject the negative root.

Example 28: Formula, fractional roots

Solve $6x^2 - 5x - 6 = 0$.

Solution:

1. $D = 25 + 144 = 169 = 13^2$.
2. $x = \dfrac{5 \pm 13}{12} = \dfrac{18}{12} = \dfrac{3}{2}$ or $\dfrac{-8}{12} = -\dfrac{2}{3}$.

Final Answer: $x = \dfrac{3}{2}, -\dfrac{2}{3}$.

Takeaway: Perfect-square $D$ ⇒ rational roots even with $a \neq 1$.

Example 29: Show real and equal for a parameter

Show that $(b - c)x^2 + (c - a)x + (a - b) = 0$ has equal roots if $2b = a + c$.

Solution:

1. Note $x = 1$ satisfies it: $(b-c) + (c-a) + (a-b) = 0$. So 1 is always a root.
2. Product of roots $= \dfrac{a - b}{b - c}$. For equal roots both are 1, so product $= 1$: $a - b = b - c \Rightarrow 2b = a + c$.

Final Answer: Equal roots ⇔ $2b = a + c$.

Takeaway: Spotting an obvious root (here $x = 1$) simplifies the proof.

Example 30: Boat and stream

A boat covers 12 km upstream and 12 km downstream in 5 hours. If the stream speed is 1 km/h, find the boat's still-water speed.

Solution:

1. Let still-water speed $= x$. $\dfrac{12}{x - 1} + \dfrac{12}{x + 1} = 5$.
2. $12(x+1) + 12(x-1) = 5(x^2 - 1) \Rightarrow 24x = 5x^2 - 5 \Rightarrow 5x^2 - 24x - 5 = 0$.
3. $(5x + 1)(x - 5) = 0 \Rightarrow x = 5$ (reject $-\tfrac{1}{5}$).

Final Answer: Still-water speed $= 5$ km/h.

Takeaway: Total time (sum of two journeys) gives the quadratic here.

Example 31: Mixed — find $m$ for equal roots

Find $m$ so that $(m + 1)x^2 + 2(m + 3)x + (m + 8) = 0$ has equal roots.

Solution:

1. $D = [2(m+3)]^2 - 4(m+1)(m+8) = 0$.
2. $4(m^2 + 6m + 9) - 4(m^2 + 9m + 8) = 0 \Rightarrow 4(6m + 9 - 9m - 8) = 0 \Rightarrow -3m + 1 = 0$.
3. $m = \dfrac{1}{3}$.

Final Answer: $m = \dfrac{1}{3}$.

Takeaway: Expand $D = 0$ carefully; the $m^2$ terms cancel, leaving a linear equation.