6 Quadratic Equation

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 $-3$.

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 ($D$): $$D = b^2 - 4ac = (-7)^2 - 4(2)(3)$$ $$D = 49 - 24 = 25$$

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: $x = \frac{7-5}{4} = \frac{2}{4} = \frac{1}{2}$

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.
$$D = b^2 - 4ac = 0$$

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. $$D = 4^2 - 4(1)(-4) = 16 + 16 = 32$$

$$x = \frac{-4 \pm \sqrt{32}}{2}$$ 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: $$(Base)^2 + (Altitude)^2 = (Hypotenuse)^2$$ $$x^2 + (x - 7)^2 = 13^2$$

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.

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