Application of Quadratic Equation

Quadratic equations are among the most useful mathematical tools for solving real-life problems. They arise when a quantity depends on the square of an unknown value, such as the area of a rectangle, the diagonal of a field, the time taken during a journey, or the relationship between speed and distance.

\[ \begin{array}{|l|l|} \hline \textbf{Concept} & \textbf{Formula} \\ \hline \text{Standard quadratic equation} & ax^{2} + bx + c = 0 \\[4pt] \hline \text{Quadratic formula} & x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \\[8pt] \hline \text{Discriminant} & D = b^{2} - 4ac \\[4pt] \hline \text{Pythagoras theorem} & \text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Height}^{2} \\[4pt] \hline \text{Speed} & \text{Speed} = \dfrac{\text{Distance}}{\text{Time}} \\[8pt] \hline \text{Time} & \text{Time} = \dfrac{\text{Distance}}{\text{Speed}} \\[8pt] \hline \end{array} \]

Question: There is a rectangular onion storehouse in the farm of Mr. Ratnakarrao at Tivasa. The length of rectangular base is more than its breadth by 7 m and diagonal is more than length by 1 m. Find length and breadth of the storehouse.

Solution:

Given Information

Let the breadth be \[ x \text{ m}. \]

Then:

Breadth \[ = x \text{ m} \]

Length \[ = x + 7 \text{ m} \]

Diagonal \[ = x + 8 \text{ m} \]

Forming the Equation

Since \[ ABC \] is a right-angled triangle, by Pythagoras theorem:

\[ (x + 8)^{2} = x^{2} + (x + 7)^{2} \]

Expanding:

\[ x^{2} + 16x + 64 = x^{2} + x^{2} + 14x + 49 \]

\[ x^{2} + 16x + 64 = 2x^{2} + 14x + 49 \]

\[ 0 = x^{2} - 2x - 15 \]

\[ x^{2} - 2x - 15 = 0 \]

Solving by Factorisation

\[ x^{2} - 5x + 3x - 15 = 0 \]

\[ x(x - 5) + 3(x - 5) = 0 \]

\[ (x - 5)(x + 3) = 0 \]

\[ x = 5 \quad \text{or} \quad x = -3 \]

Checking Validity

Since breadth cannot be negative: \[ x = -3 \text{ is rejected} \]

Therefore: \[ x = 5 \]

Final Answer

Breadth \[ = 5 \text{ m} \]

Length \[ = 5 + 7 = 12 \text{ m} \]

Diagonal \[ = 5 + 8 = 13 \text{ m} \]

Answer: The breadth of the storehouse is 5 m and the length is 12 m.

Question: A train travels 360km at a uniform speed. If the speed had been 5km/h more, it would have taken 48 minutes less to cover the same distance. Find the original speed of the train.

Solution:

Given Information

Let the original speed be \[ x \text{ km/h}. \]

Then:

\[ \begin{array}{|l|l|l|} \hline \textbf{Quantity} & \textbf{Original Journey} & \textbf{Faster Journey} \\ \hline \text{Distance} & 360 \text{ km} & 360 \text{ km} \\[4pt] \hline \text{Speed} & x \text{ km/h} & x + 5 \text{ km/h} \\[4pt] \hline \text{Time} & \dfrac{360}{x} \text{ h} & \dfrac{360}{x + 5} \text{ h} \\[8pt] \hline \end{array} \]

The time saved is 48 minutes:

\[ 48 \text{ minutes} = \frac{48}{60} = \frac{4}{5} \text{ hour} \]

Forming the Equation

\[ \frac{360}{x} - \frac{360}{x + 5} = \frac{4}{5} \]

Taking LCM:

\[ \frac{360(x + 5) - 360x}{x(x + 5)} = \frac{4}{5} \]

\[ \frac{1800}{x(x + 5)} = \frac{4}{5} \]

Cross-multiplying:

\[ 1800 \times 5 = 4x(x + 5) \]

\[ 9000 = 4x^{2} + 20x \]

\[ 4x^{2} + 20x - 9000 = 0 \]

Dividing by 4:

\[ x^{2} + 5x - 2250 = 0 \]

Solving the Equation

\[ x^{2} + 50x - 45x - 2250 = 0 \]

\[ x(x + 50) - 45(x + 50) = 0 \]

\[ (x + 50)(x - 45) = 0 \]

\[ x = -50 \quad \text{or} \quad x = 45 \]

Checking Validity

Speed cannot be negative, so: \[ x = -50 \text{ is rejected} \]

Therefore: \[ x = 45 \]

Final Answer:

The original speed of the train is 45 km/h.