Quadratic Equations: Word Problems and Nature of Roots Class 10

In algebraic mathematics, linear equations describe steady, constant relationships. But when physical quantities involve acceleration, geometric areas, or reciprocal rates of work, the relationship bends into a curve governed by a second-degree polynomial: the Quadratic Equation: $ax^2 + bx + c = 0 \quad (a \ne 0)$

In CBSE Class 10 Mathematics, Chapter 4 (Quadratic Equations) accounts for approximately $6$ to $8$ marks. Board examiners focus heavily on two distinct categories: evaluating the Nature of Roots using the discriminant ($D = b^2 - 4ac$) to find unknown parameters like $k$, and modeling complex real-world geometric and speed-distance word problems.

In this guide, we master both dimensions of quadratic analysis.

What You Will Learn

• The Discriminant ($D = b^2 - 4ac$) and the three conditions for nature of roots
• Finding the unknown constant $k$ when an equation has two equal real roots ($D = 0$)
• Geometric word problems: Right triangle hypotenuse and rectangular park dimensions
• Speed-Distance-Time word problems: The Delayed Aeroplane Problem (5-mark board classic)
• Algebraic methods: Factoring by splitting the middle term vs. the Quadratic Formula
• Rejecting extraneous negative roots

1. The Discriminant and Nature of Roots

For the general quadratic equation $ax^2 + bx + c = 0$ ($a \ne 0$), the quantity $D = b^2 - 4ac$ is called the Discriminant because it discriminates between the types of roots without needing to solve the equation:

                            The Discriminant Decision Tree
                                     D = b² - 4ac
                                          |
       +----------------------------------+----------------------------------+
       |                                  |                                  |
    D > 0                              D = 0                              D < 0
Two Distinct Real Roots            Two EQUAL Real Roots               No Real Roots
x = (-b ± √D) / 2a                 x = -b / 2a                        (Roots are imaginary)

Solved Example: Finding $k$ for Equal Real Roots (CBSE 2-Mark Classic)

Problem: Find the values of $k$ for which the quadratic equation $2x^2 + kx + 3 = 0$ has two equal real roots.

Solution:

1. Comparing with standard form $ax^2 + bx + c = 0$: $a = 2, \quad b = k, \quad c = 3$
2. The condition for two equal real roots is: $\mathbf{D = 0 \implies b^2 - 4ac = 0}$
3. Substitute the values: $k^2 - 4(2)(3) = 0$ $k^2 - 24 = 0 \implies k^2 = 24$ $\mathbf{k = \pm \sqrt{24} = \pm 2\sqrt{6}}$
4. Therefore, <u>the values of $k$ are $+2\sqrt{6}$ and $-2\sqrt{6}$</u>.

2. Geometric Word Problems

Solved Example: The Right-Angled Triangle (NCERT Classic)

Problem: The altitude of a right triangle is $7\text{ cm}$ less than its base. If the hypotenuse is $13\text{ cm}$, find the other two sides.

Solution:

1. Let the base of the right triangle be $x\text{ cm}$.
   • The altitude (height) is $7\text{ cm}$ less: $(x - 7)\text{ cm}$.
   • Hypotenuse $= 13\text{ cm}$.
2. By the Pythagoras Theorem: $\text{Base}^2 + \text{Altitude}^2 = \text{Hypotenuse}^2$ $x^2 + (x - 7)^2 = 13^2$ $x^2 + (x^2 - 14x + 49) = 169$ $2x^2 - 14x + 49 - 169 = 0$ $2x^2 - 14x - 120 = 0$
3. Divide the entire equation by $2$: $\mathbf{x^2 - 7x - 60 = 0}$
4. Factorize (Product $= -60$, Sum $= -7 \implies -12$ and $+5$): $(x - 12)(x + 5) = 0 \implies x = 12 \quad \text{or} \quad x = -5$
5. Reject Extraneous Root: A physical length cannot be negative ($x \ne -5$). Therefore, $x = \mathbf{12\text{ cm}}$.
   • Base $= 12\text{ cm}$.
   • Altitude $= 12 - 7 = \mathbf{5\text{ cm}}$.
6. Therefore, <u>the base is $12 ext{ cm}$ and the altitude is $5 ext{ cm}$</u>.

3. Speed-Distance-Time: The Delayed Aeroplane Problem (5-Mark Classic)

Problem Statement:

An aeroplane left $30\text{ minutes}$ later than the scheduled time and in order to reach its destination $1500\text{ km}$ away in time, it had to increase its speed by $250\text{ km/h}$ from its usual speed. Find its usual speed.

    Total Distance = 1500 km
    Normal Speed = x km/h                Increased Speed = (x + 250) km/h
    Normal Time T1 = 1500 / x            Fast Time T2 = 1500 / (x + 250)
    Time Difference: T1 - T2 = 30 minutes = 1/2 hour!

Step-by-Step Solution:

1. Let the usual speed of the aeroplane be $x\text{ km/h}$.
   • Increased speed $= (x + 250)\text{ km/h}$.
2. Total distance $= 1500\text{ km}$.
   • Scheduled time: $T_1 = \frac{1500}{x}\text{ hours}$.
   • Faster time: $T_2 = \frac{1500}{x + 250}\text{ hours}$.
3. Formulate the Equation: The delay was $30\text{ minutes} = \frac{30}{60} = \frac{1}{2}\text{ hour}$: $T_1 - T_2 = \frac{1}{2}$ $\frac{1500}{x} - \frac{1500}{x + 250} = \frac{1}{2}$
4. Factor Out 1500 and Simplify: $1500 \left[ \frac{(x + 250) - x}{x(x + 250)} \right] = \frac{1}{2}$ $1500 \left[ \frac{250}{x^2 + 250x} \right] = \frac{1}{2}$ $\frac{375000}{x^2 + 250x} = \frac{1}{2}$ $x^2 + 250x = 375000 \times 2$ $\mathbf{x^2 + 250x - 750000 = 0}$
5. Factorize the Large Quadratic: Look for two numbers whose product is $-750,000$ and sum is $+250$: Notice that $1000 \times (-750) = -750,000$ and $1000 - 750 = 250$! $x^2 + 1000x - 750x - 750000 = 0$ $x(x + 1000) - 750(x + 1000) = 0$ $(x + 1000)(x - 750) = 0$ $x = -1000 \quad \text{or} \quad x = 750$
6. Reject Extraneous Root: Speed cannot be negative ($x \ne -1000$).
7. Therefore, <u>the usual speed of the aeroplane is $750 ext{ km/h}$</u>.

4. Summary and Examination Tips

Exam Tip: In questions where an aeroplane or train is delayed by minutes (e.g., 30 minutes), ALWAYS convert minutes to hours (rac{30}{60} = rac{1}{2} ext{ hr}) before setting up the equation! Writing $1500/x - 1500/(x+250) = 30$ produces an answer off by a factor of 60!

Common Mistake: Forgetting the $\pm$ sign when solving $k^2 = 24$. Writing only $k = 2\sqrt{6}$ misses half the solution; $k = \pm 2\sqrt{6}$!