Math Notes For Class 10 Chapter 1 Ex 1.2 Solved & Quiz
What are the Core Formulas and Definitions for this Exercise?
In Matric Mathematics (Class 10), solving quadratic equations using the Quadratic Formula is a fundamental algebraic technique. Any quadratic equation in one variable $x$ can be written in the standard form:
$$ax^2 + bx + c = 0 \quad \text{where } a \neq 0 \text{ and } a, b, c \in \mathbb{R}$$By applying the method of completing the square to the standard quadratic equation, we derive the universal Quadratic Formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$Derivation of the Quadratic Formula
- Given $ax^2 + bx + c = 0$, divide the entire equation by $a$ ($a \neq 0$): $$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$
- Shift the constant term to the right-hand side: $$x^2 + \frac{b}{a}x = -\frac{c}{a}$$
- Add $\left(\frac{b}{2a}\right)^2$ to both sides to complete the square: $$x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}$$
- Simplify both sides: $$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2}$$
- Take the square root on both sides: $$x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$
- Isolate $x$: $$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
| Term / Component | Symbolic Representation | Role in Quadratic Formula |
|---|---|---|
| Quadratic Coefficient | $a$ | Coefficient of $x^2$; must never equal zero ($a \neq 0$). |
| Linear Coefficient | $b$ | Coefficient of $x$; determines the linear term shift. |
| Constant Term | $c$ | Independent scalar constant value. |
| Discriminant | $D = b^2 - 4ac$ | Determines the nature of roots (real, rational, irrational, or complex). |
How to Solve All Exercise Questions Step-by-Step?
Question 1: Solve the equation using the quadratic formula: $2 - x^2 = 7x$.
Solution:
Step 1: Rearrange the equation into standard quadratic form $ax^2 + bx + c = 0$:
$$-x^2 - 7x + 2 = 0 \implies x^2 + 7x - 2 = 0$$
Step 2: Identify the coefficients $a$, $b$, and $c$:
$$a = 1, \quad b = 7, \quad c = -2$$
Step 3: Substitute $a$, $b$, and $c$ into the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$$x = \frac{-(7) \pm \sqrt{(7)^2 - 4(1)(-2)}}{2(1)}$$
$$x = \frac{-7 \pm \sqrt{49 + 8}}{2}$$
$$x = \frac{-7 \pm \sqrt{57}}{2}$$
Since $57$ has no perfect square factors, the roots cannot be simplified further.
Answer: Solution Set $= \left\{ \frac{-7 \pm \sqrt{57}}{2} \right\}$
Question 2: Solve the equation using the quadratic formula: $5x^2 + 8x + 1 = 0$.
Solution:
Step 1: The equation is already in standard form $ax^2 + bx + c = 0$. Identify the coefficients:
$$a = 5, \quad b = 8, \quad c = 1$$
Step 2: Apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$$x = \frac{-(8) \pm \sqrt{(8)^2 - 4(5)(1)}}{2(5)}$$
$$x = \frac{-8 \pm \sqrt{64 - 20}}{10}$$
$$x = \frac{-8 \pm \sqrt{44}}{10}$$
Step 3: Simplify the radical expression $\sqrt{44} = \sqrt{4 \times 11} = 2\sqrt{11}$:
$$x = \frac{-8 \pm 2\sqrt{11}}{10}$$
Step 4: Factor out $2$ from the numerator and simplify with the denominator:
$$x = \frac{2(-4 \pm \sqrt{11})}{10} = \frac{-4 \pm \sqrt{11}}{5}$$
Answer: Solution Set $= \left\{ \frac{-4 \pm \sqrt{11}}{5} \right\}$
Question 3: Solve the equation using the quadratic formula: $\sqrt{3}x^2 + x = 4\sqrt{3}$.
Solution:
Step 1: Transform to standard quadratic form $ax^2 + bx + c = 0$:
$$\sqrt{3}x^2 + x - 4\sqrt{3} = 0$$
Step 2: Identify the coefficients:
$$a = \sqrt{3}, \quad b = 1, \quad c = -4\sqrt{3}$$
Step 3: Substitute into the quadratic formula:
$$x = \frac{-(1) \pm \sqrt{(1)^2 - 4(\sqrt{3})(-4\sqrt{3})}}{2(\sqrt{3})}$$
$$x = \frac{-1 \pm \sqrt{1 + 16(\sqrt{3})^2}}{2\sqrt{3}}$$
$$x = \frac{-1 \pm \sqrt{1 + 16(3)}}{2\sqrt{3}} = \frac{-1 \pm \sqrt{1 + 48}}{2\sqrt{3}}$$
$$x = \frac{-1 \pm \sqrt{49}}{2\sqrt{3}} = \frac{-1 \pm 7}{2\sqrt{3}}$$
Step 4: Separate into two distinct linear cases:
Part (i):
$$x = \frac{-1 + 7}{2\sqrt{3}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}} = \sqrt{3}$$
Part (ii):
$$x = \frac{-1 - 7}{2\sqrt{3}} = \frac{-8}{2\sqrt{3}} = -\frac{4}{\sqrt{3}}$$
Answer: Solution Set $= \left\{ \sqrt{3}, -\frac{4}{\sqrt{3}} \right\}$
Question 4: Solve the equation using the quadratic formula: $4x^2 - 14 = 3x$.
Solution:
Step 1: Rearrange into standard form $ax^2 + bx + c = 0$:
$$4x^2 - 3x - 14 = 0$$
Step 2: Identify coefficients:
$$a = 4, \quad b = -3, \quad c = -14$$
Step 3: Apply the quadratic formula:
$$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(4)(-14)}}{2(4)}$$
$$x = \frac{3 \pm \sqrt{9 + 224}}{8}$$
$$x = \frac{3 \pm \sqrt{233}}{8}$$
Since $233$ is a prime number, $\sqrt{233}$ cannot be reduced further.
Answer: Solution Set $= \left\{ \frac{3 \pm \sqrt{233}}{8} \right\}$
Question 5: Solve the equation using the quadratic formula: $6x^2 - 3 - 7x = 0$.
Solution:
Step 1: Rearrange the terms in descending order of powers of $x$:
$$6x^2 - 7x - 3 = 0$$
Step 2: Identify coefficients:
$$a = 6, \quad b = -7, \quad c = -3$$
Step 3: Apply the quadratic formula:
$$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(6)(-3)}}{2(6)}$$
$$x = \frac{7 \pm \sqrt{49 + 72}}{12}$$
$$x = \frac{7 \pm \sqrt{121}}{12}$$
$$x = \frac{7 \pm 11}{12}$$
Step 4: Evaluate the two paths:
Part (i):
$$x = \frac{7 + 11}{12} = \frac{18}{12} = \frac{3}{2}$$
Part (ii):
$$x = \frac{7 - 11}{12} = \frac{-4}{12} = -\frac{1}{3}$$
Answer: Solution Set $= \left\{ \frac{3}{2}, -\frac{1}{3} \right\}$
Question 6: Solve the equation using the quadratic formula: $3x^2 + 8x + 2 = 0$.
Solution:
Step 1: Identify coefficients directly from standard form:
$$a = 3, \quad b = 8, \quad c = 2$$
Step 2: Apply the quadratic formula:
$$x = \frac{-(8) \pm \sqrt{(8)^2 - 4(3)(2)}}{2(3)}$$
$$x = \frac{-8 \pm \sqrt{64 - 24}}{6}$$
$$x = \frac{-8 \pm \sqrt{40}}{6}$$
Step 3: Simplify the radical $\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$:
$$x = \frac{-8 \pm 2\sqrt{10}}{6}$$
Step 4: Factor out $2$ from the numerator:
$$x = \frac{2(-4 \pm \sqrt{10})}{6} = \frac{-4 \pm \sqrt{10}}{3}$$
Answer: Solution Set $= \left\{ \frac{-4 \pm \sqrt{10}}{3} \right\}$
Interactive Practice Quiz: Test Your Understanding (Clickable MCQs)
Q1: What is the standard form of a quadratic equation in one variable?
Q2: For the equation $3x^2 - 5x + 2 = 0$, what are the values of $a$, $b$, and $c$?
Q3: If the discriminant $b^2 - 4ac = 0$, the roots of the quadratic equation are:
Q4: What are the roots of $x^2 - 9 = 0$ using the quadratic formula?
Q5: What is the result of simplifying $\frac{-8 \pm \sqrt{44}}{10}$?
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