8.2 Linear Equations, Systems & Quadratic Equations

Linear Equations, Systems & Quadratic Equations

Equations assert the exact mathematical equality of two algebraic expressions. Mastering equation-solving methodologies is crucial for success on the NTS GAT General Quantitative Reasoning section. This section explores single-variable linear equations, two-variable linear systems, quadratic equations, discriminant analysis, and Vieta's formulas.

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1. Single-Variable Linear Equations

A linear equation in one variable can be expressed in standard form as $ax + b = 0$ (where $a \neq 0$). Linear equations are first-degree equations possessing exactly one unique real solution.

Standard 5-Step Solution Strategy

1. Clear Fractions and Decimals: Multiply all terms on both sides of the equation by the Least Common Denominator (LCD) of all fractions.
2. Remove Parentheses: Apply the distributive law to clear grouping symbols.
3. Isolate Variable Terms: Use addition and subtraction properties of equality to collect all terms containing the target variable on one side and all constants on the opposite side.
4. Solve for Variable: Divide both sides by the numerical coefficient of the isolated variable.
5. Verify Solution: Substitute the resulting root back into the original equation.

Step-by-Step Worked Example

$\text{Solve for } x: \quad \frac{2x - 3}{4} - \frac{x + 1}{3} = \frac{x - 2}{6}$

1. Find LCD: The LCD of 4, 3, and 6 is $12$.
2. Multiply every term by 12: $12 \cdot \left(\frac{2x - 3}{4}\right) - 12 \cdot \left(\frac{x + 1}{3}\right) = 12 \cdot \left(\frac{x - 2}{6}\right)$ $3(2x - 3) - 4(x + 1) = 2(x - 2)$
3. Distribute and combine terms: $6x - 9 - 4x - 4 = 2x - 4 \implies 2x - 13 = 2x - 4$
4. Analyze outcome: Subtracting $2x$ from both sides yields $-13 = -4$, which is a contradiction. Hence, this equation has no real solution (empty solution set $\emptyset$). (Note: If isolating variables yields a true statement like $0 = 0$, the equation is an identity with infinitely many solutions).

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2. Systems of Two-Variable Linear Equations

A system of two linear equations in two variables has the general form: $\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}$

Algebraic Solution Methods

• Substitution Method: Isolate one variable in one equation and substitute its expression into the second equation. Best suited when one coefficient is $\pm 1$.
• Elimination Method: Multiply one or both equations by non-zero constants so the coefficients of one variable become additive inverses, then add the equations to eliminate that variable.

Worked Example: Elimination Method

$\begin{cases} 4x + 3y = 25 \\ 3x - 2y = 6 \end{cases}$

1. Multiply the first equation by 2 and the second equation by 3 to eliminate $y$: $\begin{cases} 8x + 6y = 50 \\ 9x - 6y = 18 \end{cases}$
2. Add the two equations: $17x = 68 \implies x = 4$
3. Substitute $x = 4$ into the second equation: $3(4) - 2y = 6 \implies 12 - 2y = 6 \implies -2y = -6 \implies y = 3$
4. Solution Set: $(x, y) = (4, 3)$.

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3. Quadratic Equations

A quadratic equation is a second-degree equation in standard form: $ax^2 + bx + c = 0 \quad (a \neq 0)$

Primary Solution Techniques

A. Factoring (Zero-Product Property)

If $ax^2 + bx + c = (px + q)(rx + s) = 0$, set $px + q = 0$ or $rx + s = 0$. Example: $x^2 - 7x + 12 = 0 \implies (x - 3)(x - 4) = 0 \implies x = 3 \text{ or } x = 4$.

B. The Quadratic Formula

For any quadratic equation $ax^2 + bx + c = 0$, the roots are given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Worked Example: Quadratic Formula

$\text{Solve: } 2x^2 - 5x - 3 = 0$

Here $a = 2, b = -5, c = -3$. $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm \sqrt{49}}{4} = \frac{5 \pm 7}{4}$

• $x_1 = \frac{5 + 7}{4} = \frac{12}{4} = 3$
• $x_2 = \frac{5 - 7}{4} = \frac{-2}{4} = -\frac{1}{2}$

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4. Discriminant Analysis

The quantity under the radical sign in the quadratic formula is the discriminant, denoted by $\Delta = b^2 - 4ac$.

$\Delta = b^2 - 4ac$

• $\Delta > 0$: Two distinct real roots.
  • If $\Delta$ is a perfect square, roots are rational.
  • If $\Delta$ is not a perfect square, roots are irrational conjugate pairs ($p \pm \sqrt{q}$).
• $\Delta = 0$: Exactly one real repeated root (double root), given by $x = -\frac{b}{2a}$.
• $\Delta < 0$: Two non-real complex conjugate roots ($p \pm iq$).

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5. Vieta's Formulas & Quantitative Shortcuts

Vieta's Formulas establish direct connections between polynomial coefficients and root sums/products. For $ax^2 + bx + c = 0$ with roots $x_1, x_2$:

$\text{Sum of Roots: } x_1 + x_2 = -\frac{b}{a}$ $\text{Product of Roots: } x_1 \cdot x_2 = \frac{c}{a}$

Derived Quantitative Shortcuts

1. Sum of Squares of Roots: $x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1x_2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2 - 2ac}{a^2}$
2. Sum of Reciprocals of Roots: $\frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2}{x_1 x_2} = \frac{-b/a}{c/a} = -\frac{b}{c}$
3. Difference of Roots: $|x_1 - x_2| = \frac{\sqrt{b^2 - 4ac}}{|a|} = \frac{\sqrt{\Delta}}{|a|}$

Worked Example: Vieta Shortcut

$\text{If } \alpha, \beta \text{ are roots of } 3x^2 - 12x + 5 = 0, \text{ compute } \alpha^2 + \beta^2.$

1. Find sum and product: $\alpha + \beta = -\frac{-12}{3} = 4$ and $\alpha\beta = \frac{5}{3}$.
2. Compute $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (4)^2 - 2\left(\frac{5}{3}\right) = 16 - \frac{10}{3} = \frac{38}{3}$.