Two Quadratic Equation Questions for Class 10

This lesson covers two common Class 10 quadratic equation questions: proving a condition for equal roots, and a speed word problem solved with a quadratic.

Question 1: Condition for equal roots

We are given the quadratic equation

$(1 + m^2)x^2 + 2mcx + (c^2 - a^2) = 0$

and told it has two equal roots. We must prove that $c^2 = a^2(1 + m^2)$.

Identify the coefficients

Comparing with $ax^2 + bx + c = 0$:

$A = 1 + m^2, \quad B = 2mc, \quad C = c^2 - a^2$

Compute the discriminant

$B^2 - 4AC = (2mc)^2 - 4(1 + m^2)(c^2 - a^2)$

Expanding each part:

$(2mc)^2 = 4m^2c^2$

$4(1 + m^2)(c^2 - a^2) = 4c^2 - 4a^2 + 4m^2c^2 - 4m^2a^2$

So

$B^2 - 4AC = 4m^2c^2 - 4c^2 + 4a^2 - 4m^2c^2 + 4m^2a^2 = -4c^2 + 4a^2 + 4m^2a^2$

Apply the equal-roots condition

For equal roots the discriminant is zero:

$-4c^2 + 4a^2 + 4m^2a^2 = 0$

Dividing by $4$ and rearranging:

$a^2 + m^2a^2 = c^2$

$c^2 = a^2(1 + m^2)$

which is what we needed to prove.

Question 2: Finding the train's speed

A passenger train takes $3$ hours less for a journey of $360$ km when its speed is increased by $10$ km/h from its usual speed. Find the actual speed.

Set up the variables

Let the usual speed be $x$ km/h. Using $\text{time} = \dfrac{\text{distance}}{\text{speed}}$:

$\text{usual time} = \frac{360}{x}, \qquad \text{new time} = \frac{360}{x + 10}$

The faster trip takes $3$ hours less, so

$\frac{360}{x} - \frac{360}{x + 10} = 3$

Form the quadratic

Taking $360$ out and combining the fractions:

$360\left(\frac{1}{x} - \frac{1}{x + 10}\right) = 360 \cdot \frac{(x + 10) - x}{x(x + 10)} = \frac{3600}{x(x + 10)} = 3$

So $3600 = 3x(x + 10)$, giving $1200 = x^2 + 10x$, that is

$x^2 + 10x - 1200 = 0$

Solve by factorisation

We need two numbers with product $-1200$ and sum $10$: these are $40$ and $-30$.

$(x + 40)(x - 30) = 0$

$x = -40 \quad \text{or} \quad x = 30$

Speed cannot be negative, so we reject $x = -40$. The actual speed is

$x = 30 \text{ km/h}$

Key takeaways

• A quadratic has equal roots exactly when its discriminant $b^2 - 4ac$ is zero; setting that to zero gives the required relationship between constants.
• Expand and simplify the discriminant carefully so terms cancel cleanly.
• Word problems on time, speed, and distance often reduce to a quadratic; solve by factorisation and discard any root that is physically impossible (here a negative speed).