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Math Notes For Class 10 Chapter 2 Ex 2.1 Solved & Quiz
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Math Notes For Class 10 Chapter 2 Ex 2.1 Solved & Quiz
In the study of quadratic equations of the standard form $ax^2 + bx + c = 0$ (where $a \neq 0$), the expression $b^2 - 4ac$ is defined as the Discriminant ($\text{Disc}$). The discriminant determines the fundamental nature of the roots without explicitly solving the equation.
| Discriminant Value ($\text{Disc} = b^2 - 4ac$) | Nature of Roots | algebraic Characteristic |
|---|---|---|
| $b^2 - 4ac > 0$ (Perfect Square) | Real, Rational, and Unequal | Two distinct rational numbers |
| $b^2 - 4ac > 0$ (Not a Perfect Square) | Real, Irrational, and Unequal | Surds in conjugate pairs |
| $b^2 - 4ac = 0$ | Real, Rational, and Equal | Coincident roots ($x = -\frac{b}{2a}$) |
| $b^2 - 4ac < 0$ | Imaginary / Complex Conjugates | Contains $i = \sqrt{-1}$ |
Question 1 (Part i): Find the discriminant of the quadratic equation $2x^2 + 3x - 1 = 0$.
Solution:
Comparing $2x^2 + 3x - 1 = 0$ with $ax^2 + bx + c = 0$, we have:
$a = 2, \quad b = 3, \quad c = -1$
Using the discriminant formula:
$$\text{Disc} = b^2 - 4ac$$
$$\text{Disc} = (3)^2 - 4(2)(-1) = 9 + 8 = 17$$
Answer: $$\text{Disc} = 17$$
Question 1 (Part ii): Find the discriminant of the quadratic equation $6x^2 - 8x + 3 = 0$.
Solution:
Comparing with standard form $ax^2 + bx + c = 0$:
$a = 6, \quad b = -8, \quad c = 3$
Substituting into the formula:
$$\text{Disc} = (-8)^2 - 4(6)(3)$$
$$\text{Disc} = 64 - 72 = -8$$
Answer: $$\text{Disc} = -8$$
Question 1 (Part iii): Find the discriminant of the quadratic equation $9x^2 - 24x + 16 = 0$.
Solution:
Here coefficients are:
$a = 9, \quad b = -24, \quad c = 16$
$$\text{Disc} = (-24)^2 - 4(9)(16)$$
$$\text{Disc} = 576 - 576 = 0$$
Answer: $$\text{Disc} = 0$$
Question 1 (Part iv): Find the discriminant of the quadratic equation $4x^2 - 7x - 2 = 0$.
Solution:
Comparing coefficients:
$a = 4, \quad b = -7, \quad c = -2$
$$\text{Disc} = (-7)^2 - 4(4)(-2)$$
$$\text{Disc} = 49 + 32 = 81$$
Answer: $$\text{Disc} = 81$$
Question 2 (Part i): Find the nature of roots of $x^2 - 23x + 120 = 0$ and verify the result by solving the equation.
Solution:
$a = 1, \quad b = -23, \quad c = 120$
$$\text{Disc} = (-23)^2 - 4(1)(120) = 529 - 480 = 49 = (7)^2$$
Since $\text{Disc} > 0$ and is a perfect square, the roots are real, rational, and unequal.
Verification:
By quadratic formula: $x = \frac{-b \pm \sqrt{\text{Disc}}}{2a}$
$$x = \frac{-(-23) \pm \sqrt{49}}{2(1)} = \frac{23 \pm 7}{2}$$
$$x_1 = \frac{23 + 7}{2} = 15, \quad x_2 = \frac{23 - 7}{2} = 8$$
The roots $15$ and $8$ are real, rational, and unequal. Verified!
Answer: $$\text{Nature: Real, Rational, Unequal}; \quad \text{Roots} = \{15, 8\}$$
Question 2 (Part ii): Find the nature of roots of $2x^2 + 3x + 7 = 0$ and verify the result by solving the equation.
Solution:
$a = 2, \quad b = 3, \quad c = 7$
$$\text{Disc} = (3)^2 - 4(2)(7) = 9 - 56 = -47$$
Since $\text{Disc} < 0$, the roots are imaginary (complex conjugates) and unequal.
Verification:
$$x = \frac{-3 \pm \sqrt{-47}}{2(2)} = \frac{-3 \pm i\sqrt{47}}{4}$$
Since the solutions contain $i = \sqrt{-1}$, the roots are imaginary. Verified!
Answer: $$\text{Nature: Imaginary (Complex Conjugates)}; \quad x = \frac{-3 \pm i\sqrt{47}}{4}$$
Question 2 (Part iii): Find the nature of roots of $16x^2 - 24x + 9 = 0$ and verify the result by solving the equation.
Solution:
$a = 16, \quad b = -24, \quad c = 9$
$$\text{Disc} = (-24)^2 - 4(16)(9) = 576 - 576 = 0$$
Since $\text{Disc} = 0$, the roots are real, rational, and equal.
Verification:
$$(4x - 3)^2 = 0 \implies 4x - 3 = 0 \implies x = \frac{3}{4}$$
Both roots are $x = \frac{3}{4}, \frac{3}{4}$, which are rational and equal. Verified!
Answer: $$\text{Nature: Real, Rational, Equal}; \quad x = \frac{3}{4}$$
Question 2 (Part iv): Find the nature of roots of $3x^2 + 7x - 13 = 0$ and verify the result by solving the equation.
Solution:
$a = 3, \quad b = 7, \quad c = -13$
$$\text{Disc} = (7)^2 - 4(3)(-13) = 49 + 156 = 205$$
Since $\text{Disc} > 0$ and $205$ is NOT a perfect square, the roots are real, irrational, and unequal.
Verification:
$$x = \frac{-7 \pm \sqrt{205}}{2(3)} = \frac{-7 \pm \sqrt{205}}{6}$$
Since $\sqrt{205}$ is an irrational number, the roots are irrational and unequal. Verified!
Answer: $$\text{Nature: Real, Irrational, Unequal}; \quad x = \frac{-7 \pm \sqrt{205}}{6}$$
Question 3: For what value of $k$, the expression $k^2 x^2 + 2(k+1)x + 4$ is a perfect square?
Solution:
A quadratic expression $Ax^2 + Bx + C$ is a perfect square if its discriminant equals zero ($\text{Disc} = 0$).
Here $A = k^2, \quad B = 2(k+1), \quad C = 4$.
$$\text{Disc} = B