Section 1: Determining the Nature of Roots
Note: For any quadratic equation $ax^2 + bx + c = 0$, nature depends on Discriminant ($D = b^2 - 4ac$).
Solution: Here, $a = 2, b = 7, c = 3$.
$$\text{Discriminant } (D) = b^2 - 4ac = (7)^2 - 4 \cdot 2 \cdot 3$$
$$D = 49 - 24 = 25$$
Since $25 > 0$ (Positive and a perfect square), the roots are real, rational, and unequal.
Solution: Here, $a = 3, b = -2\sqrt{6}, c = 2$.
$$D = b^2 - 4ac = (-2\sqrt{6})^2 - 4 \cdot 3 \cdot 2$$
$$D = (4 \times 6) - 24 = 24 - 24 = 0$$
Since $D = 0$, the roots are real and equal.
Solution: Here, $a = 2, b = -7, c = 9$.
$$D = b^2 - 4ac = (-7)^2 - 4 \cdot 2 \cdot 9$$
$$D = 49 - 72 = -23$$
Since $D < 0$ (Negative), there are no real roots.
Section 2: Finding Values of $k$ for Real and Equal Roots ($D = 0$)
Solution: Here $a = 3, b = -5, c = 2k$. Set $b^2 - 4ac = 0$:
$$\implies (-5)^2 - 4 \cdot 3 \cdot (2k) = 0$$
$$\implies 25 - 24k = 0 \implies 24k = 25 \implies k = \frac{25}{24}$$
Solution: Here $a = 9, b = -24, c = k$. Set $b^2 - 4ac = 0$:
$$\implies (-24)^2 - 4 \cdot 9 \cdot k = 0$$
$$\implies 576 - 36k = 0 \implies 36k = 576 \implies k = \frac{576}{36} = 16$$
Solution: Here $a = 1, b = -2(5+2k), c = 3(7+10k)$. Since roots are equal, $b^2 - 4ac = 0$:
$$\implies \{-2(5 + 2k)\}^2 - 4 \cdot 1 \cdot 3(7 + 10k) = 0$$
$$\implies 4(25 + 20k + 4k^2) - 12(7 + 10k) = 0$$
Divide by 4 to simplify:
$$\implies (25 + 20k + 4k^2) - 3(7 + 10k) = 0$$
$$\implies 25 + 20k + 4k^2 - 21 - 30k = 0$$
$$\implies 4k^2 - 10k + 4 = 0$$
Divide by 2:
$$\implies 2k^2 - 5k + 2 = 0$$
$$\implies 2k^2 - 4k - k + 2 = 0 \implies 2k(k - 2) - 1(k - 2) = 0$$
$$\implies (k - 2)(2k - 1) = 0 \implies k = 2 \text{ or } k = \frac{1}{2}$$
Section 3: Forming Quadratic Equations from Given Roots
Formula: $x^2 - (\text{Sum of Roots})x + (\text{Product of Roots}) = 0$
• Sum $= 4 + 2 = 6$
• Product $= 4 \times 2 = 8$
Equation: $x^2 - 6x + 8 = 0$
• Sum $= (-4) + (-3) = -7$
• Product $= (-4) \times (-3) = 12$
Equation: $x^2 - (-7)x + 12 = 0 \implies x^2 + 7x + 12 = 0$
Section 4: Proofs and Algebraic Derivations
Proof: Since the roots are equal, Discriminant $= 0$.
$$\implies \{-2(ac + bd)\}^2 - 4(a^2 + b^2)(c^2 + d^2) = 0$$
$$\implies 4(a^2c^2 + b^2d^2 + 2abcd) - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0$$
Divide by 4 and expand terms:
$$\implies a^2c^2 + b^2d^2 + 2abcd - a^2c^2 - a^2d^2 - b^2c^2 - b^2d^2 = 0$$
Canceling out identical terms ($a^2c^2$ and $b^2d^2$):
$$\implies 2abcd - a^2d^2 - b^2c^2 = 0$$
Multiply by $-1$ to rearrange:
$$\implies a^2d^2 - 2abcd + b^2c^2 = 0$$
This fits the algebraic formula $(A-B)^2 = A^2 - 2AB + B^2$:
$$\implies (ad - bc)^2 = 0 \implies ad - bc = 0 \implies ad = bc$$
Dividing both sides by $bd$ or restructuring gives:
$$\frac{a}{c} = \frac{b}{d} \quad \text{(Proved)}$$
Proof: Let us evaluate the discriminant ($D$):
$$D = \{2(a + b)\}^2 - 4 \cdot [2(a^2 + b^2)] \cdot 1$$
$$D = 4(a^2 + b^2 + 2ab) - 8a^2 - 8b^2$$
$$D = 4a^2 + 4b^2 + 8ab - 8a^2 - 8b^2$$
$$D = -4a^2 - 4b^2 + 8ab = -4(a^2 - 2ab + b^2)$$
$$D = -4(a - b)^2$$
Since a square term $(a-b)^2$ is always positive for any real distinct values ($a \neq b$), multiplying it by $-4$ ensures that $D$ is strictly negative ($D < 0$). Thus, the roots cannot be real.
Section 5: Evaluations using Symmetric Roots ($\alpha, \beta$)
From the equation, Sum ($\alpha + \beta$) $= -\frac{b}{a} = -\frac{2}{5}$ and Product ($\alpha\beta$) $= \frac{c}{a} = -\frac{3}{5}$.
1. Evaluate $\alpha^2 + \beta^2$:
$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{2}{5}\right)^2 - 2\left(-\frac{3}{5}\right)$$
$$= \frac{4}{25} + \frac{6}{5} = \frac{4 + 30}{25} = \frac{34}{25}$$
2. Evaluate $\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}$:
$$\frac{\alpha^3 + \beta^3}{\alpha\beta} = \frac{(\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)}{\alpha\beta}$$
Numerator ($\alpha^3+\beta^3$) $= \left(-\frac{2}{5}\right)^3 - 3\left(-\frac{3}{5}\right)\left(-\frac{2}{5}\right) = -\frac{8}{125} - \frac{18}{25} = \frac{-8 - 90}{125} = -\frac{98}{125}$
Dividing by $\alpha\beta$:
$$\text{Value} = \left(-\frac{98}{125}\right) \div \left(-\frac{3}{5}\right) = \frac{-98}{125} \times \frac{-5}{3} = \frac{98}{75}$$
Solution: Let the roots be $\alpha$ and $2\alpha$.
$$\text{Sum of roots: } \alpha + 2\alpha = -\frac{b}{a} \implies 3\alpha = -\frac{b}{a} \implies \alpha = -\frac{b}{3a}$$
$$\text{Product of roots: } \alpha \cdot (2\alpha) = \frac{c}{a} \implies 2\alpha^2 = \frac{c}{a}$$
Substitute the value of $\alpha$ into the product equation:
$$2\left(-\frac{b}{3a}\right)^2 = \frac{c}{a} \implies 2\left(\frac{b^2}{9a^2}\right) = \frac{c}{a}$$
$$\implies \frac{2b^2}{9a^2} = \frac{c}{a}$$
Multiply both sides by $9a^2$:
$$2b^2 = 9ac \quad \text{(Proved)}$$
Section 6: Multiple Choice Questions (MCQs)
- (a) 2
- (b) -2
- (c) 6
- (d) -6
Explanation: $\text{Sum} = -\frac{b}{a} = -\frac{-6}{1} = 6$.
- (a) -2
- (b) -8
- (c) 8
- (d) 12
Explanation: Write in standard form: $x^2 - 3x + (k - 10) = 0$. Product $= c/a = k - 10 = -2 \implies k = 10 - 2 = 8$.
Section 7: Short Answer Questions (SA)
Solution:
$$\text{Sum of roots} = -\frac{2}{k}, \quad \text{Product of roots} = \frac{3k}{k} = 3$$
Given that they are equal:
$$-\frac{2}{k} = 3 \implies 3k = -2 \implies k = -\frac{2}{3}$$
Solution: Here, $\alpha + \beta = 22$ and $\alpha\beta = 105$.
Using identity: $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$
$$(\alpha - \beta)^2 = (22)^2 - 4(105)$$
$$(\alpha - \beta)^2 = 484 - 420 = 64$$
$$\implies \alpha - \beta = \pm\sqrt{64} = \pm 8$$