Algebraic Mastery for PPSC

The Nature of Quadratic Roots

In competitive exams, speed is everything. You don't always need to find the exact roots of a quadratic equation $ax^2 + bx + c = 0$ . Often, you just need to know what kind of roots it has: are they real, imaginary, or equal? This is where the discriminant comes in.

\Delta = b^2 - 4ac

The value of the discriminant tells you everything you need to know about the roots without solving for them. There are three possibilities to remember.

Discriminant Value	Nature of Roots
$\Delta > 0$	Real and distinct (unequal)
$\Delta = 0$	Real and equal (one repeated root)
$\Delta < 0$	Complex or imaginary (not real)

Let's try an example. What is the nature of the roots for the equation $2x^2 - 7x + 6 = 0$?

Here, $a=2$, $b=-7$, and $c=6$. We calculate the discriminant:

\Delta = (-7)^2 - 4(2)(6) = 49 - 48 = 1

Since $\Delta = 1$ , which is greater than zero, the equation has two distinct real roots. No need to solve the whole quadratic formula; you've answered the question in seconds.

Polynomials and Their Factors

When dealing with higher-degree polynomials like $x^3 - 2x^2 - 5x + 6$ , finding roots can be tedious. The Remainder and Factor Theorems provide powerful shortcuts.

The Remainder Theorem states that if you divide a polynomial $P(x)$ by a linear factor $(x-a)$ , the remainder is simply $P(a)$ .

The Factor Theorem is a direct consequence of this. It states that $(x-a)$ is a factor of $P(x)$ if and only if $P(a) = 0$ . In other words, if plugging in a value makes the polynomial equal zero, you've found a root.

Is $(x-1)$ a factor of $P(x) = x^3 - 2x^2 - 5x + 6$ ?

According to the Factor Theorem, we just need to check if $P(1) = 0$ . $P(1) = (1)^3 - 2(1)^2 - 5(1) + 6 = 1 - 2 - 5 + 6 = 0$ . Yes, it is. This means $x=1$ is a root.

Once you find one factor, you can simplify the problem. For high-degree polynomials, this is where synthetic division becomes an indispensable tool. It's a faster, cleaner way to perform polynomial division, especially with linear divisors.

Let's divide $x^3 - 2x^2 - 5x + 6$ by $(x-1)$ using synthetic division.

Root: 1
Coefficients: 1, -2, -5, 6

1 | 1  -2  -5   6
  |    1  -1  -6
  ------------------
    1  -1  -6   0 --> Remainder

Resulting coefficients: 1, -1, -6
This corresponds to the quadratic: $x^2 - x - 6$.

Now we just have to solve the simpler quadratic $x^2 - x - 6 = 0$ , which factors into $(x-3)(x+2) = 0$ . So, the other roots are $x=3$ and $x=-2$ . The full set of roots for the original cubic polynomial is $\{-2, 1, 3\}$ .

Solving Systems of Equations

Systems of linear equations are common in aptitude tests. While substitution works, it can be slow and error-prone with three or more variables. Cramer's Rule offers a systematic approach using determinants.

Consider a system of two linear equations:

\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}

First, we find the determinant of the main coefficient matrix, D.

D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1b_2 - a_2b_1

Next, we find $D_x$ by replacing the x-column coefficients with the constants, and $D_y$ by replacing the y-column coefficients with the constants.

D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} \quad \text{and} \quad D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix}

The solution is then straightforward, provided $D \neq 0$ .

x = \frac{D_x}{D} \quad \text{and} \quad y = \frac{D_y}{D}

Another powerful shortcut involves the relationship between a polynomial's roots and its coefficients, described by Vieta's formulas for a quadratic equation $ax^2+bx+c=0$ with roots $\alpha$ and $\beta$ :

\alpha + \beta = -\frac{b}{a} \\ \alpha \beta = \frac{c}{a}

This is incredibly useful in MCQs. For example, if a question asks for the sum of the roots of $3x^2 - 5x - 2 = 0$, you don't need to find the roots. You can just calculate:

Sum of roots = $-(-5)/3 = 5/3$.

Algebra in Word Problems

Algebraic modelling is crucial for solving word problems related to age, work, and rates, which are staples of the General Ability portion of recruitment tests.

The key is to translate the words into mathematical equations. Let's look at a classic work-rate problem.

Ali can paint a house in 6 days, and Bilal can do it in 12 days. How long will it take them to paint the house if they work together?

First, determine their individual rates of work. The rate is the fraction of the job completed per unit of time.

Ali's rate: 1/6 of the house per day.
Bilal's rate: 1/12 of the house per day.

When they work together, their rates add up. Let $T$ be the time it takes them together.

\text{Combined Rate} = \frac{1}{6} + \frac{1}{12}

To add these fractions, we find a common denominator, which is 12.

\frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}

If they complete 1/4 of the job each day, the entire job (1 whole job) will take the reciprocal of that rate. So, $T = 4$ days.

This structured approach of defining variables, setting up rates, and forming an equation is the most reliable way to solve these problems.

Quiz Questions 1/6

What is the nature of the roots for the quadratic equation $3x^2 - 5x + 3 = 0$ ?

Quiz Questions 2/6

Without solving, what is the product of the roots of the equation $2x^2 + 8x - 5 = 0$ ?

Mastering PPSC EST Mathematics