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Quadratic Equations: Competency Practice

Highway underpass clearance, speed-distance modeling, discriminant parameters, and digit systems.

7 problems·20–25 min·★★★★★
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  1. Problem 1
    quadratic-equations·★★★★★
    A parabolic highway underpass arch has height $y$ (in metres) at horizontal distance $x$ (in metres) from the road centerline modeled by $y = 8 - \frac{1}{8} x^2$. (i) What is the shape of the curve represented by the quadratic polynomial? (ii) Find the height of the arch at a distance of $4\text{ m}$ from the centerline. (iii) A cargo truck of width $8\text{ m}$ and height $5.5\text{ m}$ approaches along the centerline; determine mathematically whether the truck can pass through the arch without touching the ceiling.
    ▶Answer
    (i) Parabola (downward-opening), (ii) $6\text{ m}$, (iii) Yes, the vehicle will pass safely
    ▶Step-by-step solution
    1. (i) The graph of any quadratic polynomial $p(x) = ax^2 + bx + c$ with negative leading coefficient ($-\frac{1}{8} < 0$) is a downward-opening parabola.
    2. (ii) Height at $x = 4\text{ m}$: $y(4) = 8 - \frac{1}{8}(4)^2 = 6\text{ m}$.
    3. (iii) A truck of width $8\text{ m}$ extends to $x = \pm 4\text{ m}$. Arch clearance at the edge is $y(\pm 4) = 8 - \frac{1}{8}(4)^2 = 6\text{ m}$, which is > the truck height of $5.5\text{ m}$. Thus yes, the vehicle will pass safely.
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  2. Problem 2
    quadratic-equations·★★★★★
    A two-digit number is $4$ times the sum of its digits and twice the product of its digits. Find the number.
    ▶Answer
    $36$
    ▶Step-by-step solution
    1. Let the tens digit be $x$ and units digit be $y$. The number is $10x + y$.
    2. Condition 1: $10x + y = 4(x + y) \implies 6x = 3y \implies y = 2x$.
    3. Condition 2: $10x + y = 2xy$. Substituting $y = 2x$ gives $12x = 4x^2 \implies 4x(x - 3) = 0$.
    4. Since $x \ne 0$, $x = 3$ and $y = 6$.
    5. Therefore, the original number is $10(3) + 6 = 36$.
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  3. Problem 3
    Nature of Roots·★★★★☆
    For which values of $k$ does $x^2-kx+36=0$ have equal roots?
    1. (A)$k=36$
    2. (B)$k=12$ only
    3. (C)$k=-12$ only
    4. (D)$k=-12$ or $k=12$
    ▶Answer
    $k=-12$ or $k=12$ (option D).
    ▶Step-by-step solution
    1. Equal roots require the discriminant to be zero.
    2. Here, $D=(-k)^2-4(36)=k^2-144$.
    3. Solving $D=0$ gives $k=-12$ or $k=12$.
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  4. Problem 4
    Nature of Roots·★★★★☆
    Which equation has two distinct real roots?
    1. (A)$3x^2 + 24x + 48 = 0$
    2. (B)$3x^2 + 3x - 6 = 0$
    3. (C)$x^2 + 4x + 4 = 0$
    4. (D)$3x^2 + 2x + 3 = 0$
    ▶Answer
    Option B: $3x^2 + 3x - 6 = 0$
    ▶Step-by-step solution
    1. Compute $D=b^2-4ac$ for each option.
    2. For option B, $D=81$.
    3. This discriminant corresponds to two distinct real roots.
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  5. Problem 5
    Quadratic Formula·★★★★☆
    Solve $2x^2 - 8x + 3 = 0$ using the quadratic formula.
    ▶Answer
    The roots are $x = \frac{4 - \sqrt{10}}{2}$ or $x = \frac{4 + \sqrt{10}}{2}$.
    ▶Step-by-step solution
    1. Use $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ with $a=2$, $b=-8$, and $c=3$.
    2. The discriminant is $D=40$.
    3. Hence, $x = \frac{4 - \sqrt{10}}{2}$ or $x = \frac{4 + \sqrt{10}}{2}$.
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  6. Problem 6
    Quadratic Word Problems·★★★★★
    A train covers 150 km at a constant speed. At 15 km/h faster, it would take 20 minutes less. Find its original speed.
    ▶Answer
    The original speed is $75$ km/h.
    ▶Step-by-step solution
    1. Let the original speed be $v$ km/h. The time difference is $150/v-150/(v+15)=\frac{1}{3}$ hour.
    2. Clearing denominators gives a quadratic whose roots are $75$ and $-90$.
    3. A speed must be positive, so the original speed is $75$ km/h.
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  7. Problem 7
    Quadratic Word Problems·★★★★☆
    A parent's present age is 3 more than the square of the child's present age. When the child reaches the parent's present age, the parent will be 5 less than 22 times the child's present age. Find both present ages.
    ▶Answer
    The child is $11$ years old and the parent is $124$ years old.
    ▶Step-by-step solution
    1. Let the child's age be $c$; then the parent's age is $c^2+3$.
    2. When the child reaches that age, the parent's age will be $2(c^2+3)-c$. Use the stated comparison to form the quadratic.
    3. The valid age is $c=11$, giving parent age $11^2+3=124$ years.
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