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Math Notes For Class 11 Chapter 3 Ex 3.1 Solved & Quiz
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Math Notes For Class 11 Chapter 3 Ex 3.1 Solved & Quiz
A quadratic equation in one variable $x$ is an algebraic equation of second degree, written in standard form as:
$$ax^2 + bx + c = 0 \quad \text{where } a, b, c \in \mathbb{R} \text{ and } a \neq 0$$To solve quadratic equations and evaluate their properties, three primary analytical methods are used in Class 11 Mathematics: Factorization, Completing the Square, and the Quadratic Formula.
| Condition of Discriminant ($\Delta = b^2 - 4ac$) | Nature of Roots | Graphical Representation |
|---|---|---|
| $\Delta > 0$ and a perfect square | Real, Rational, and Unequal (Distinct) | Cuts x-axis at two rational points |
| $\Delta > 0$ and not a perfect square | Real, Irrational, and Unequal (Distinct) | Cuts x-axis at two irrational points |
| $\Delta = 0$ | Real, Rational, and Equal (Repeated) | Touches x-axis at exactly one point |
| $\Delta < 0$ | Complex Conjugates / Imaginary | Does not intersect the x-axis |
Question 1 (Part i): Solve the quadratic equation by factorization: $x^2 + 7x + 12 = 0$.
Solution:
We need two numbers whose product is $12$ and whose sum is $7$. These numbers are $3$ and $4$.
$$x^2 + 4x + 3x + 12 = 0$$
Factor by grouping terms:
$$x(x + 4) + 3(x + 4) = 0$$
$$(x + 4)(x + 3) = 0$$
Setting each factor equal to zero:
$$x + 4 = 0 \implies x = -4$$
$$x + 3 = 0 \implies x = -3$$
Answer: $$\text{Solution Set} = \{-4, -3\}$$
Question 1 (Part ii): Solve the quadratic equation by factorization: $2x^2 - 5x - 3 = 0$.
Solution:
Product required: $2 \times (-3) = -6$, Sum required: $-5$. The factors are $-6$ and $+1$.
$$2x^2 - 6x + x - 3 = 0$$
Factor by grouping:
$$2x(x - 3) + 1(x - 3) = 0$$
$$(2x + 1)(x - 3) = 0$$
Equating factors to zero:
$$2x + 1 = 0 \implies x = -\frac{1}{2}$$
$$x - 3 = 0 \implies x = 3$$
Answer: $$\text{Solution Set} = \left\{-\frac{1}{2}, 3\right\}$$
Question 2 (Part i): Solve by completing the square: $x^2 - 6x + 5 = 0$.
Solution:
Move constant term to the right-hand side:
$$x^2 - 6x = -5$$
Take half of the coefficient of $x$, which is $\frac{-6}{2} = -3$, square it to get $9$, and add to both sides:
$$x^2 - 6x + 9 = -5 + 9$$
$$(x - 3)^2 = 4$$
Taking square root on both sides:
$$x - 3 = \pm \sqrt{4}$$
$$x - 3 = \pm 2$$
This splits into two cases:
$$x = 3 + 2 = 5 \quad \text{or} \quad x = 3 - 2 = 1$$
Answer: $$\text{Solution Set} = \{1, 5\}$$
Question 2 (Part ii): Solve by completing the square: $2x^2 + 12x - 14 = 0$.
Solution:
Divide the entire equation by the leading coefficient $2$:
$$x^2 + 6x - 7 = 0$$
Shift constant term to RHS:
$$x^2 + 6x = 7$$
Add $\left(\frac{6}{2}\right)^2 = 9$ to both sides:
$$x^2 + 6x + 9 = 7 + 9$$
$$(x + 3)^2 = 16$$
Taking square root on both sides:
$$x + 3 = \pm 4$$
$$x = -3 + 4 = 1 \quad \text{or} \quad x = -3 - 4 = -7$$
Answer: $$\text{Solution Set} = \{-7, 1\}$$
Question 3 (Part i): Solve using the quadratic formula: $3x^2 - 5x + 2 = 0$.
Solution:
Comparing $3x^2 - 5x + 2 = 0$ with $ax^2 + bx + c = 0$, we have $a = 3$, $b = -5$, $c = 2$.
Apply quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
$$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(2)}}{2(3)}$$
$$x = \frac{5 \pm \sqrt{25 - 24}}{6} = \frac{5 \pm \sqrt{1}}{6} = \frac{5 \pm 1}{6}$$
Thus:
$$x = \frac{5 + 1}{6} = 1 \quad \text{or} \quad x = \frac{5 - 1}{6} = \frac{4}{6} = \frac{2}{3}$$
Answer: $$\text{Solution Set} = \left\{\frac{2}{3}, 1\right\}$$
Question 3 (Part ii): Solve using the quadratic formula: $x^2 + x + 1 = 0$.
Solution:
Here $a = 1$, $b = 1$, $c = 1$.
$$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)}$$
$$x = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2}$$
Since $\sqrt{-3} = i\sqrt{3}$:
$$x = \frac{-1 \pm i\sqrt{3}}{2}$$
Answer: $$\text{Solution Set} = \left\{\frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}\right\}$$
Question 4 (Part i): Determine the nature of the roots of $x^2 - 4x + 4 = 0$.
Solution:
Comparing with standard form: $a = 1$, $b = -4$, $c = 4$.
Calculate Discriminant $\Delta$:
$$\Delta = b^2 - 4ac = (-4)^2 - 4(1)(4) = 16 - 16 = 0$$
Since $\Delta = 0$, the roots are real, rational, and equal.
Answer: $$\text{Roots are Real, Rational, and Equal}$$
Question 4 (Part ii): Determine the nature of the roots of $2x^2 + 3x + 5 = 0$.
Solution:
Comparing with standard form: $a = 2$, $b = 3$, $c = 5$.
Calculate Discriminant $\Delta$:
$$\Delta = b^2 - 4ac = (3)^2 - 4(2)(5) = 9 - 40 = -31$$
Since $\Delta < 0$, the discriminant is negative.
Answer: $$\text{Roots are Complex Conjugates (Imaginary)}$$
Question 5: Find the value of $k$ for which the roots of $(k+1)x^2 + 2(k+3)x + (k+8) = 0$ are equal.
Solution:
Here $a = k+1$, $b = 2(k+3)$, $c = k+8$.
For the equation to have equal roots, the discriminant must be zero ($\Delta = 0$):
$$\Delta = b^2 - 4ac = 0$$
$$[2(k+3)]^2 - 4(k+1)(k+8) = 0$$
$$4(k^2 + 6k + 9) - 4(k^2 + 9k + 8) = 0$$
Divide the entire equation by $4$:
$$(k^2 + 6k + 9) - (k^2 + 9k + 8) = 0$$
$$k^2 + 6k + 9 - k^2 - 9k - 8 = 0$$
$$-3k + 1 = 0 \implies 3k = 1 \implies k = \frac{1}{3}$$
Answer: $$k = \frac{1}{3}$$
Question 6: Show that the roots of the equation $(a^2 - bc)x^2 + 2(b^2 - ca)x + (c^2 - ab) = 0$ are equal if $a^3 + b^3 + c^3 = 3abc$ or $b = 0$.
Solution:
Let $A = a^2 - bc$, $B = 2(b^2 - ca)$, and $C = c^2 - ab$.
The roots are equal if and only if Discriminant $\Delta = B^2 - 4AC = 0$.
$$\Delta = [2(b^2 - ca)]^2 - 4(a^2 - bc)(c^2 - ab)$$
$$\Delta = 4(b^2 - ca)^2 - 4(a^2 - bc)(c^2 - ab)$$
Divide by $4$ and expand:
$$\Delta / 4 = (b^4 - 2ab^2c + c^2a^2) - (a^2c^2 - a^3b - bc^3 + ab^2c)$$
$$\Delta / 4 = b^4 - 2ab^2c + a^2c^2 - a^2c^2 + a^3b + bc^3 - ab^2c$$
$$\Delta / 4 = b^4 - 3ab^2c + a^3b + bc^3$$
Factor out $b$:
$$\Delta / 4 = b(a^3 + b^3 + c^3 - 3abc)$$
Setting $\Delta = 0$ gives:
$$b(a^3 + b^3 + c^3 - 3abc) = 0$$
This implies either:
$$b = 0 \quad \text{or} \quad a^3 + b^3 + c^3 = 3abc$$
Answer: $$\text{Proved that roots are equal when } b = 0 \text{ or } a^3 + b^3 + c^3 = 3abc.$$
Q1: What is the discriminant of the quadratic equation $3x^2 - 2x + 1 = 0$?