Part 1: Basic Concepts of Quadratic Equations
A Quadratic Equation is a polynomial equation of the second degree. The standard form is:
$$ax^2 + bx + c = 0$$
Where $a, b, c$ are real numbers and $a \neq 0$.
Key Methods to Solve
- Factorization: Splitting the middle term.
- Quadratic Formula: Used when factorization is difficult.
The Quadratic Formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Here, the term $D = b^2 - 4ac$ is called the Discriminant.
- If $D > 0$, roots are real and distinct.
- If $D = 0$, roots are real and equal.
- If $D < 0$, roots are imaginary (no real solution).
Part 2: 6 Solved Problems
Problem 1: Solve by Factorization
Solve the equation for $x$: $$x^2 - 5x + 6 = 0$$
Solution:
We need two numbers that multiply to $+6$ and add up to $-5$. These numbers are $-2$ and $-3$.
We need two numbers that multiply to $+6$ and add up to $-5$. These numbers are $-2$ and $-3$.
Split the middle term:
$$x^2 - 2x - 3x + 6 = 0$$
Group terms:
$$x(x - 2) - 3(x - 2) = 0$$
$$(x - 2)(x - 3) = 0$$
Answer: $x = 2$ or $x = 3$
Problem 2: Using the Quadratic Formula
Find the roots of: $$2x^2 - 7x + 3 = 0$$
Solution:
Identify coefficients: $a = 2, b = -7, c = 3$.
Calculate the discriminant ($D$): $$D = b^2 - 4ac = (-7)^2 - 4(2)(3)$$ $$D = 49 - 24 = 25$$
Identify coefficients: $a = 2, b = -7, c = 3$.
Calculate the discriminant ($D$): $$D = b^2 - 4ac = (-7)^2 - 4(2)(3)$$ $$D = 49 - 24 = 25$$
Apply formula:
$$x = \frac{-(-7) \pm \sqrt{25}}{2(2)}$$
$$x = \frac{7 \pm 5}{4}$$
Case 1: $x = \frac{7+5}{4} = \frac{12}{4} = 3$
Case 2: $x = \frac{7-5}{4} = \frac{2}{4} = \frac{1}{2}$
Case 2: $x = \frac{7-5}{4} = \frac{2}{4} = \frac{1}{2}$
Answer: $x = 3, \frac{1}{2}$
Problem 3: Nature of Roots
Find the value of $k$ for which the quadratic equation $2x^2 + kx + 3 = 0$ has two real equal roots.
Solution:
For equal roots, the Discriminant ($D$) must be zero.
$$D = b^2 - 4ac = 0$$
For equal roots, the Discriminant ($D$) must be zero.
$$D = b^2 - 4ac = 0$$
Substitute values ($a=2, b=k, c=3$):
$$k^2 - 4(2)(3) = 0$$
$$k^2 - 24 = 0$$
$$k^2 = 24$$
Answer: $k = \pm \sqrt{24} = \pm 2\sqrt{6}$
Problem 4: Roots involving Irrational Numbers
Solve for $x$: $$x^2 + 4x - 4 = 0$$
Solution:
Here $a=1, b=4, c=-4$. Use the formula. $$D = 4^2 - 4(1)(-4) = 16 + 16 = 32$$
Here $a=1, b=4, c=-4$. Use the formula. $$D = 4^2 - 4(1)(-4) = 16 + 16 = 32$$
$$x = \frac{-4 \pm \sqrt{32}}{2}$$
Simplify $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$.
$$x = \frac{-4 \pm 4\sqrt{2}}{2}$$
Answer: $x = -2 \pm 2\sqrt{2}$
Problem 5: Geometry Word Problem
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:
Let the base be $x$ cm.
Then, altitude = $(x - 7)$ cm.
By Pythagoras theorem: $$(Base)^2 + (Altitude)^2 = (Hypotenuse)^2$$ $$x^2 + (x - 7)^2 = 13^2$$
Let the base be $x$ cm.
Then, altitude = $(x - 7)$ cm.
By Pythagoras theorem: $$(Base)^2 + (Altitude)^2 = (Hypotenuse)^2$$ $$x^2 + (x - 7)^2 = 13^2$$
Expand and simplify:
$$x^2 + (x^2 - 14x + 49) = 169$$
$$2x^2 - 14x + 49 - 169 = 0$$
$$2x^2 - 14x - 120 = 0$$
Divide by 2:
$$x^2 - 7x - 60 = 0$$
Factorize ($12$ and $-5$):
$$(x - 12)(x + 5) = 0$$
$x = 12$ or $x = -5$. Since length cannot be negative, $x = 12$.
Answer: Base = 12 cm, Altitude = 5 cm
Problem 6: Speed and Distance Word Problem
A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Find the speed of the train.
Solution:
Let the usual speed be $x$ km/h.
Time taken usually = $\frac{480}{x}$ hours.
New speed = $(x - 8)$ km/h.
New time = $\frac{480}{x - 8}$ hours.
Let the usual speed be $x$ km/h.
Time taken usually = $\frac{480}{x}$ hours.
New speed = $(x - 8)$ km/h.
New time = $\frac{480}{x - 8}$ hours.
According to the question, the difference in time is 3 hours:
$$\frac{480}{x - 8} - \frac{480}{x} = 3$$
Solve for $x$:
$$480 \left[ \frac{x - (x - 8)}{x(x - 8)} \right] = 3$$
$$480 \left[ \frac{8}{x^2 - 8x} \right] = 3$$
$$3840 = 3(x^2 - 8x)$$
$$1280 = x^2 - 8x$$
$$x^2 - 8x - 1280 = 0$$
Using factorization ($x^2 - 40x + 32x - 1280 = 0$):
$$(x - 40)(x + 32) = 0$$
$x = 40$ or $x = -32$. Speed cannot be negative.
Answer: Speed of the train = 40 km/h