Math Notes For Class 10 Chapter 1 Ex 1.1 Solved & Quiz
What are the Core Formulas and Definitions for this Exercise?
A quadratic equation in one variable $x$ is an equation that can be written in the form $ax^2 + bx + c = 0$, where $a \neq 0$ and $a, b, c \in \mathbb{R}$. This form is known as the standard form or general form of a quadratic equation. If $b = 0$, the equation reduces to $ax^2 + c = 0$, which is termed a pure quadratic equation.
To solve quadratic equations in Exercise 1.1, two primary analytical methods are used:
- Factorization Method: Expressing the quadratic expression $ax^2 + bx + c$ as a product of two linear factors $(px + q)(rx + s) = 0$ and applying the Zero-Factor Property ($A \cdot B = 0 \implies A = 0$ or $B = 0$).
- Completing the Square Method: Manipulating the algebraic terms to transform the equation into the form $(x + k)^2 = d$, allowing direct extraction of square roots.
| Concept / Method | Mathematical Representation | Key Structural Requirement |
|---|---|---|
| Standard Quadratic Form | $ax^2 + bx + c = 0$ | $a \neq 0$, arrangement in descending powers of $x$. |
| Pure Quadratic Form | $ax^2 + c = 0$ | Coefficient of linear term $b = 0$. |
| Zero-Factor Property | If $a \cdot b = 0$, then $a = 0$ or $b = 0$ | Used in factorization to split factors into linear equations. |
| Completing Square Constant | $\left(\frac{b}{2a}\right)^2$ | Term added to both sides after making leading coefficient equal to $1$. |
How to Solve All Exercise Questions Step-by-Step?
Question 1 (Part i): Write the equation $(x + 7)(x - 3) = -7$ in standard form $ax^2 + bx + c = 0$ and point out pure quadratic equations.
Solution:
Given equation:
$$(x + 7)(x - 3) = -7$$
Expand the left-hand side:
$$x(x - 3) + 7(x - 3) = -7$$
$$x^2 - 3x + 7x - 21 = -7$$
Combine like terms:
$$x^2 + 4x - 21 = -7$$
Add $7$ to both sides:
$$x^2 + 4x - 21 + 7 = 0$$
$$x^2 + 4x - 14 = 0$$
This is of the form $ax^2 + bx + c = 0$ with $a = 1$, $b = 4$, and $c = -14$. Since $b \neq 0$, it is a standard quadratic equation.
Answer: $$x^2 + 4x - 14 = 0 \quad \text{(Standard Quadratic Equation)}$$
Question 1 (Part ii): Write $\frac{x^2 + 4}{3} - \frac{x}{7} = 1$ in standard form and point out pure quadratic equations.
Solution:
Given equation:
$$\frac{x^2 + 4}{3} - \frac{x}{7} = 1$$
Multiply the entire equation by the LCM of denominators ($3 \times 7 = 21$):
$$21 \cdot \left(\frac{x^2 + 4}{3}\right) - 21 \cdot \left(\frac{x}{7}\right) = 21 \cdot 1$$
$$7(x^2 + 4) - 3(x) = 21$$
$$7x^2 + 28 - 3x = 21$$
Rearrange terms and subtract $21$ from both sides:
$$7x^2 - 3x + 28 - 21 = 0$$
$$7x^2 - 3x + 7 = 0$$
This is in standard quadratic form $ax^2 + bx + c = 0$.
Answer: $$7x^2 - 3x + 7 = 0 \quad \text{(Standard Quadratic Equation)}$$
Question 1 (Part iii): Write $\frac{x}{x+1} + \frac{x+1}{x} = 6$ in standard form and point out pure quadratic equations.
Solution:
Given equation:
$$\frac{x}{x+1} + \frac{x+1}{x} = 6$$
Multiply both sides by the common denominator $x(x + 1)$:
$$x(x) + (x + 1)(x + 1) = 6x(x + 1)$$
$$x^2 + (x^2 + 2x + 1) = 6x^2 + 6x$$
$$2x^2 + 2x + 1 = 6x^2 + 6x$$
Rearrange all terms to the right side:
$$6x^2 - 2x^2 + 6x - 2x - 1 = 0$$
$$4x^2 + 4x - 1 = 0$$
This is in standard quadratic form $ax^2 + bx + c = 0$.
Answer: $$4x^2 + 4x - 1 = 0 \quad \text{(Standard Quadratic Equation)}$$
Question 2 (Part i): Solve by factorization: $x^2 - x - 20 = 0$.
Solution:
Given equation:
$$x^2 - x - 20 = 0$$
Find two numbers whose product is $-20$ and sum is $-1$. These numbers are $-5$ and $+4$.
$$x^2 - 5x + 4x - 20 = 0$$
Group terms:
$$x(x - 5) + 4(x - 5) = 0$$
$$(x - 5)(x + 4) = 0$$
Apply Zero-Factor Property:
$$x - 5 = 0 \implies x = 5$$
$$x + 4 = 0 \implies x = -4$$
Answer: $$\text{Solution Set} = \{-4, 5\}$$
Question 2 (Part ii): Solve by factorization: $3y^2 = y(y - 5)$.
Solution:
Given equation:
$$3y^2 = y^2 - 5y$$
Move all terms to the left side:
$$3y^2 - y^2 + 5y = 0$$
$$2y^2 + 5y = 0$$
Factor out common term $y$:
$$y(2y + 5) = 0$$
Apply Zero-Factor Property:
$$y = 0$$
$$2y + 5 = 0 \implies 2y = -5 \implies y = -\frac{5}{2}$$
Answer: $$\text{Solution Set} = \left\{0, -\frac{5}{2}\right\}$$
Question 2 (Part iii): Solve by factorization: $4 - 32x = 17x^2$.
Solution:
Rearrange into standard form $ax^2 + bx + c = 0$:
$$17x^2 + 32x - 4 = 0$$
Product of $a$ and $c$: $17 \times (-4) = -68$. Sum required: $+32$.
The required factors are $+34$ and $-2$:
$$17x^2 + 34x - 2x - 4 = 0$$
Factor by grouping:
$$17x(x + 2) - 2(x + 2) = 0$$
$$(17x - 2)(x + 2) = 0$$
Set factors to zero:
$$17x - 2 = 0 \implies x = \frac{2}{17}$$
$$x + 2 = 0 \implies x = -2$$
Answer: $$\text{Solution Set} = \left\{-2, \frac{2}{17}\right\}$$
Question 3 (Part i): Solve by completing the square: $7x^2 + 2x - 1 = 0$.
Solution:
Step 1: Divide by leading coefficient $7$:
$$x^2 + \frac{2}{7}x - \frac{1}{7} = 0$$
Step 2: Shift constant to right side:
$$x^2 + \frac{2}{7}x = \frac{1}{7}$$
Step 3: Add $\left(\frac{1}{2} \cdot \frac{2}{7}\right)^2 = \left(\frac{1}{7}\right)^2 = \frac{1}{49}$ to both sides:
$$x^2 + \frac{2}{7}x + \left(\frac{1}{7}\right)^2 = \frac{1}{7} + \frac{1}{49}$$
$$\left(x + \frac{1}{7}\right)^2 = \frac{7 + 1}{49} = \frac{8}{49}$$
Step 4: Take square root of both sides:
$$x + \frac{1}{7} = \pm \sqrt{\frac{8}{49}} = \pm \frac{2\sqrt{2}}{7}$$
$$x = -\frac{1}{7} \pm \frac{2\sqrt{2}}{7} = \frac{-1 \pm 2\sqrt{2}}{7}$$
Answer: $$\text{Solution Set} = \left\{\frac{-1 \pm 2\sqrt{2}}{7}\right\}$$
Question 3 (Part ii): Solve by completing the square: $ax^2 + 4x - a = 0, \quad a \neq 0$.
Solution:
Divide through by $a$:
$$x^2 + \frac{4}{a}x - 1 = 0$$
Shift constant term:
$$x^2 + \frac{4}{a}x = 1$$
Add $\left(\frac{1}{2} \cdot \frac{4}{a}\right)^2 = \left(\frac{2}{a}\right)^2 = \frac{4}{a^2}$ to both sides:
$$x^2 + \frac{4}{a}x + \left(\frac{2}{a}\right)^2 = 1 + \frac{4}{a^2}$$
$$\left(x + \frac{2}{a}\right)^2 = \frac{a^2 + 4}{a^2}$$
Taking square root on both sides
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