05. Quadratic Equations in One Variable Previous Year Questions

Discriminant D = b² − 4ac = (√7)² − 4(3)(2) = 7 − 24 = −17.
Since D < 0, the equation has no real roots. [1 Mark]

a = 2, b = −5, c = −4.
x = [−b ± √(b² − 4ac)] / 2a = [5 ± √((-5)² − 4(2)(−4))] / 4 = [5 ± √(25 + 32)] / 4 = [5 ± √57] / 4. [2 Marks]
From tables: √57 ≈ 7.550.
x1 = (5 + 7.550) / 4 = 12.550 / 4 = 3.1375 ⇒ 3.14 (3 sig figs). [1 Mark]
x2 = (5 − 7.550) / 4 = −2.550 / 4 = −0.6375 ⇒ −0.638 (or −0.637). [1 Mark]

Let original speed = x km/hr. Reduced speed = (x − 20) km/hr.
Time difference: 350 / (x − 20) − 350 / x = 2. [1.5 Marks]
350 [ x − (x − 20) ] / [ x(x − 20) ] = 2 ⇒ 350(20) = 2(x² − 20x) ⇒ x² − 20x − 3500 = 0.
(x − 70)(x + 50) = 0 ⇒ Since speed > 0, x = 70 km/hr.
(a) Original speed = 70 km/hr. [1.5 Marks]
(b) Reduced speed = 70 − 20 = 50 km/hr. Time taken = 350 / 50 = 7 hours. [1 Mark]

Correct Option: (b) q² = 4pr
Discriminant D = B² − 4AC = 0 for real and equal roots.
(−q)² − 4(p)(r) = 0 ⇒ q² = 4pr. [1 Mark]

a = 2, b = −10, c = 5. Discriminant D = (−10)² − 4(2)(5) = 100 − 40 = 60.
√60 = 7.7460. [1 Mark]
x = [10 ± 7.746] / 4
x₁ = 17.746 / 4 = 4.4365 ≈ 4.44 (3 sig fig). [1 Mark]
x₂ = 2.254 / 4 = 0.5635 ≈ 0.564 (3 sig fig). [1 Mark]

Since 3 is a root of the equation, it must satisfy $x^2 - px + 3 = 0$:
$3^2 - p(3) + 3 = 0 \implies 9 - 3p + 3 = 0 \implies 12 - 3p = 0 \implies 3p = 12 \implies p = 4$.
Correct Option: (a) [1 Mark]

Step 1 Comparing with $ax^2 + bx + c = 0$: $a = 1, b = 4, c = -8$.
Discriminant $D = b^2 - 4ac = 4^2 - 4(1)(-8) = 16 + 32 = 48$.1 Mark
Step 2 Applying quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:
$x = \frac{-4 \pm \sqrt{48}}{2(1)} = \frac{-4 \pm 4\sqrt{3}}{2} = -2 \pm 2\sqrt{3}$.1 Mark
Step 3 Taking $\sqrt{3} \approx 1.732$ (or $\sqrt{48} \approx 6.928$):
$x_1 = -2 + 2(1.732) = -2 + 3.464 = 1.464 \approx \mathbf{1.5}$1 Mark
$x_2 = -2 - 2(1.732) = -2 - 3.464 = -5.464 \approx \mathbf{-5.5}$
Answers: $x = 1.5$ or $x = -5.5$.1 Mark

Step 1 Set up time expressions:
Initial time $t_1 = \frac{100}{x}\text{ hours}$.
Increased speed time $t_2 = \frac{100}{x + 5}\text{ hours}$.1 Mark

Step 2 According to given condition $t_1 - t_2 = 1$:
$\frac{100}{x} - \frac{100}{x + 5} = 1$
$100\left(\frac{x + 5 - x}{x(x + 5)}\right) = 1 \implies \frac{500}{x^2 + 5x} = 1$
$x^2 + 5x - 500 = 0$.1 Mark

Step 3 Solve quadratic equation:
$(x + 25)(x - 20) = 0 \implies x = 20$ or $x = -25$.
Since speed cannot be negative, $x = 20$.
Original speed = 20 km/hr.1 Mark

Step 1 Comparing with $ax^2 + bx + c = 0$: $a = 1, b = -7, c = 3$.
Discriminant $D = b^2 - 4ac = (-7)^2 - 4(1)(3) = 49 - 12 = 37$. [1 Mark]
Step 2 Applying quadratic formula: $x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-7) \pm \sqrt{37}}{2(1)} = \frac{7 \pm 6.083}{2}$. [1 Mark]
Step 3 Evaluating roots:
$x_1 = \frac{7 + 6.083}{2} = \frac{13.083}{2} = 6.5415 \approx \mathbf{6.54}$
$x_2 = \frac{7 - 6.083}{2} = \frac{0.917}{2} = 0.4585 \approx \mathbf{0.46}$
Answer: $x = 6.54$ or $0.46$. [1 Mark]

Step 1 Let the smaller natural number be $x$. Then the larger number is $x + 7$.
Product equation: $x(x + 7) = 450 \implies x^2 + 7x - 450 = 0$. [1 Mark]
Step 2 Factorising $x^2 + 7x - 450 = 0$:
Product $= -450$, Sum $= 7 \implies +25$ and $-18$.
$x^2 + 25x - 18x - 450 = 0 \implies x(x + 25) - 18(x + 25) = 0 \implies (x + 25)(x - 18) = 0$. [1 Mark]
Step 3 Since $x$ is a natural number ($x \in \mathbb{N}$), $x = 18$ ($x = -25$ rejected).
Larger number $= 18 + 7 = 25$.
Answer: The numbers are 18 and 25. [1 Mark]

Step 1 Identify coefficients and evaluate discriminant:
$a = 1, b = -4, c = -8$.
$D = b^2 - 4ac = (-4)^2 - 4(1)(-8) = 16 + 32 = 48$. [1 Mark]
Step 2 Apply quadratic formula:
$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-4) \pm \sqrt{48}}{2(1)} = \frac{4 \pm 4\sqrt{3}}{2} = 2 \pm 2\sqrt{3}$.
Using $\sqrt{3} \approx 1.73205 \implies 2\sqrt{3} \approx 3.4641$. [1 Mark]
Step 3 Calculate both roots to 3 significant figures:
$x_1 = 2 + 3.4641 = 5.4641 \approx \mathbf{5.46}$
$x_2 = 2 - 3.4641 = -1.4641 \approx \mathbf{-1.46}$
Answer: $x = 5.46$ or $-1.46$. [1 Mark]

Step 1 Let the two consecutive natural numbers which are multiples of 3 be $3x$ and $3x + 3 = 3(x + 1)$, where $x \in \mathbb{N}$. [1 Mark]
Step 2 Set up quadratic equation from their product:
$(3x)(3x + 3) = 810$
$9x(x + 1) = 810 \implies x(x + 1) = \frac{810}{9} = 90$
$x^2 + x - 90 = 0$. [1 Mark]
Step 3 Factorise the quadratic equation:
$x^2 + 10x - 9x - 90 = 0 \implies (x + 10)(x - 9) = 0$
$\implies x = 9$ or $x = -10$.
Since $x$ must be a natural number ($x \in \mathbb{N}$), reject $x = -10$.
$\therefore x = 9$.
First number $= 3(9) = \mathbf{27}$.
Second number $= 27 + 3 = \mathbf{30}$.
Answer: The two numbers are 27 and 30. [1 Mark]

Solution $a = 2, b = 5, c = c$.
$D = b^2 - 4ac = 5^2 - 4(2)(c) = 25 - 8c$.
Given $D = 1 \implies 25 - 8c = 1 \implies 8c = 24 \implies \mathbf{c = 3}$.
Correct Answer: (d) 3 [1 Mark]

Step 1 Let the original speed of the car be $x\text{ km/h}$.
Original time $t_1 = \frac{400}{x}\text{ hours}$.
Increased speed $= (x + 12)\text{ km/h}$.
New time $t_2 = \frac{400}{x + 12}\text{ hours}$.
Time difference $= 1\text{ hr } 40\text{ min} = 1 + \frac{40}{60} = 1 + \frac{2}{3} = \frac{5}{3}\text{ hours}$. [1 Mark]
Step 2 Formulate and solve quadratic equation:
$\frac{400}{x} - \frac{400}{x + 12} = \frac{5}{3} \implies 400 \left[\frac{x + 12 - x}{x(x + 12)}\right] = \frac{5}{3}$
$\frac{4800}{x^2 + 12x} = \frac{5}{3} \implies 5(x^2 + 12x) = 14400 \implies x^2 + 12x - 2880 = 0$. [1.5 Marks]
Step 3 Factorise quadratic equation:
$(x + 60)(x - 48) = 0 \implies x = 48 \text{ or } x = -60$.
Since speed cannot be negative, $x = 48\text{ km/h}$.
(a) $\mathbf{\text{Original speed} = 48\text{ km/h}}$. [1 Mark]
Step 4 (b) Time taken with increased speed ($48 + 12 = 60\text{ km/h}$):
$\text{Time} = \frac{400}{60} = \frac{20}{3}\text{ hours} = \mathbf{6\text{ hours } 40\text{ minutes}}$. [0.5 Mark]

Step 1 Identify coefficients and evaluate discriminant:
$a = 1, b = -6, c = -8$.
$D = b^2 - 4ac = (-6)^2 - 4(1)(-8) = 36 + 32 = 68$. [1 Mark]
Step 2 Apply quadratic formula:
$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-6) \pm \sqrt{68}}{2(1)} = \frac{6 \pm 8.2462}{2} = 3 \pm 4.1231$. [1 Mark]
Step 3 Calculate roots to 3 significant figures:
$x_1 = 3 + 4.1231 = 7.1231 \approx \mathbf{7.12}$
$x_2 = 3 - 4.1231 = -1.1231 \approx \mathbf{-1.12}$
Answer: $x = 7.12$ or $-1.12$. [1 Mark]

Step 1 Standard quadratic form: $x^2 + 7x - 7 = 0$.
Here $a = 1, b = 7, c = -7$. [1 Mark]
Step 2 Quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{7^2 - 4(1)(-7)}}{2(1)}$$ $$x = \frac{-7 \pm \sqrt{49 + 28}}{2} = \frac{-7 \pm \sqrt{77}}{2}$$ [1.5 Marks]
Step 3 Calculate decimal roots with $\sqrt{77} \approx 8.775$:
$$x_1 = \frac{-7 + 8.775}{2} = \frac{1.775}{2} = 0.8875 \approx \mathbf{0.89}$$ [0.75 Mark]
$$x_2 = \frac{-7 - 8.775}{2} = \frac{-15.775}{2} = -7.8875 \approx \mathbf{-7.89}$$ [0.75 Mark]
Answer: $x = 0.89$ or $x = -7.89$.

Step 1 For equal roots, discriminant $D = b^2 - 4ac = 0$.
Here $a = 1, b = 4k, c = k^2 - k + 2$. [1 Mark]
Step 2 Form the equation in $k$:
$$(4k)^2 - 4(1)(k^2 - k + 2) = 0$$ $$16k^2 - 4k^2 + 4k - 8 = 0 \implies 12k^2 + 4k - 8 = 0$$ [1 Mark]
Step 3 Divide by 4 and factorise:
$$3k^2 + k - 2 = 0 \implies 3k^2 + 3k - 2k - 2 = 0$$ $$3k(k + 1) - 2(k + 1) = 0 \implies (3k - 2)(k + 1) = 0$$ $$\mathbf{k = \frac{2}{3}} \quad \text{or} \quad \mathbf{k = -1}$$ [1 Mark]
Answer: $k = \frac{2}{3}$ or $k = -1$.

Step 1 Let the original number of children be $x$.
Original share of each child $= \frac{7500}{x}$.
New share of each child with $(x - 20)$ children $= \frac{7500}{x - 20}$. [1 Mark]
Step 2 Form the quadratic equation:
$$\frac{7500}{x - 20} - \frac{7500}{x} = 100$$ Divide by 100: $\frac{75}{x - 20} - \frac{75}{x} = 1$
$$75 \left(\frac{x - (x - 20)}{x(x - 20)}\right) = 1 \implies \frac{1500}{x^2 - 20x} = 1$$ $$x^2 - 20x - 1500 = 0$$ [1 Mark]
Step 3 Factorise:
$$(x - 50)(x + 30) = 0$$ Since the number of children cannot be negative, $x = 50$. [1 Mark]
Answer: The original number of children was 50.

Step 1 Compare with $ax^2 + bx + c = 0$: $a = 4, b = -5, c = -3$. [1 Mark]
Step 2 Apply quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(4)(-3)}}{2(4)}$$ $$x = \frac{5 \pm \sqrt{25 + 48}}{8} = \frac{5 \pm \sqrt{73}}{8}$$ [1.5 Marks]
Step 3 Using $\sqrt{73} \approx 8.544$:
$$x_1 = \frac{5 + 8.544}{8} = \frac{13.544}{8} = 1.693 \approx \mathbf{1.69}$$ [0.75 Mark]
$$x_2 = \frac{5 - 8.544}{8} = \frac{-3.544}{8} = -0.443 \approx \mathbf{-0.44}$$ [0.75 Mark]
Answer: $x = 1.69$ or $x = -0.44$.

Step 1 Let Vivek's age be $x$ years. Then Amit's age $= 47 - x$ years. [1 Mark]
Step 2 Formulate quadratic equation:
$$x(47 - x) = 550 \implies 47x - x^2 = 550$$ $$x^2 - 47x + 550 = 0$$ [1 Mark]
Step 3 Factorise ($25 \times 22 = 550$ and $25 + 22 = 47$):
$$(x - 25)(x - 22) = 0$$ [1 Mark]
Since Vivek is the older brother, $x = 25$, and Amit $= 47 - 25 = 22$. [1 Mark]
Answer: Vivek is 25 years old and Amit is 22 years old.