Class 10 Math Note Class 10 Chapter 2 Theory of Quadratic Equations Exercise 2.1 Solved - New 2025-2026 Syllabus Solved PDF Notes
In the study of quadratic equations of the standard general form $ax^2 + bx + c = 0$ where $a \neq 0$ and $a, b, c \in \mathbb{R}$, the nature of the roots depends entirely on the expression under the radical sign in the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. This fundamental expression $b^2 - 4ac$ is defined as the Discriminant of the quadratic equation, denoted by $\text{Disc}$ or $\Delta$.
According to the NEW 2025-2026 Single National Curriculum (SNC) textbook for Class 10 Mathematics, the value and algebraic sign of the discriminant determine the specific characteristics (nature) of the roots without explicitly solving the equation. The key rules governing the nature of roots are classified into four primary mathematical cases:
Below is the structured formula reference table essential for solving Exercise 2.1 and mastering the theory of quadratic equations for board examinations:
| Mathematical Entity / Condition | Formula / Criterion | Nature of Roots / Mathematical Result |
|---|---|---|
| Standard Quadratic Equation | $ax^2 + bx + c = 0 \quad (a \neq 0)$ | Base equation with real coefficients $a, b, c$. |
| Discriminant ($\text{Disc}$) | $\text{Disc} = \Delta = b^2 - 4ac$ | Determines the nature of roots. |
| $\Delta > 0$ and Perfect Square | $b^2 - 4ac = k^2 \quad (k \in \mathbb{Q})$ | Real, Rational, and Unequal |
| $\Delta > 0$ and Non-Perfect Square | $b^2 - 4ac \neq k^2$ | Real, Irrational, and Unequal |
| $\Delta = 0$ | $b^2 - 4ac = 0$ | Real, Rational, and Equal (Perfect Square) |
| $\Delta < 0$ | $b^2 - 4ac < 0$ | Imaginary / Complex Conjugates |
| Condition for Perfect Square Expression | $b^2 - 4ac = 0$ | The quadratic expression forms $(px + q)^2$. |
Question 1: Find the discriminant of the following given quadratic equations.
(i) $2x^2 + 3x - 1 = 0$
Solution:
Comparing with standard equation $ax^2 + bx + c = 0$:
$a = 2$, $b = 3$, $c = -1$
$$\text{Discriminant} = b^2 - 4ac$$
$$\text{Disc} = (3)^2 - 4(2)(-1)$$
$$\text{Disc} = 9 + 8 = 17$$
Answer: The discriminant is $17$.
(ii) $6x^2 - 8x + 3 = 0$
Solution:
Comparing with standard equation $ax^2 + bx + c = 0$:
$a = 6$, $b = -8$, $c = 3$
$$\text{Disc} = b^2 - 4ac$$
$$\text{Disc} = (-8)^2 - 4(6)(3)$$
$$\text{Disc} = 64 - 72 = -8$$
Answer: The discriminant is $-8$.
(iii) $9x^2 - 24x + 16 = 0$
Solution:
Comparing with standard equation $ax^2 + bx + c = 0$:
$a = 9$, $b = -24$, $c = 16$
$$\text{Disc} = b^2 - 4ac$$
$$\text{Disc} = (-24)^2 - 4(9)(16)$$
$$\text{Disc} = 576 - 576 = 0$$
Answer: The discriminant is $0$.
(iv) $4x^2 - 7x - 2 = 0$
Solution:
Comparing with standard equation $ax^2 + bx + c = 0$:
$a = 4$, $b = -7$, $c = -2$
$$\text{Disc} = b^2 - 4ac$$
$$\text{Disc} = (-7)^2 - 4(4)(-2)$$
$$\text{Disc} = 49 + 32 = 81$$
Answer: The discriminant is $81$.
Question 2: Find the nature of the roots of the following given quadratic equations and verify the result by solving the equations.
(i) $x^2 - 23x + 120 = 0$
Step 1: Determine Nature using Discriminant
$a = 1$, $b = -23$, $c = 120$
$$\text{Disc} = b^2 - 4ac = (-23)^2 - 4(1)(120) = 529 - 480 = 49$$
Since $\text{Disc} = 49 > 0$ and $49 = 7^2$ (a perfect square), the roots are real, rational, and unequal.
Step 2: Verification by Solving Equation
$$x^2 - 23x + 120 = 0$$
Using factorization method ($15 \times 8 = 120$ and $15 + 8 = 23$):
$$x^2 - 15x - 8x + 120 = 0$$
$$x(x - 15) - 8(x - 15) = 0$$
$$(x - 15)(x - 8) = 0$$
$$x = 15 \quad \text{or} \quad x = 8$$
The roots are $8$ and $15$, which are rational, real, and distinct. Hence verified.
(ii) $2x^2 + 3x + 7 = 0$
Step 1: Determine Nature using Discriminant
$a = 2$, $b = 3$, $c = 7$
$$\text{Disc} = b^2 - 4ac = (3)^2 - 4(2)(7) = 9 - 56 = -47$$
Since $\text{Disc} = -47 < 0$, the roots are imaginary (complex conjugates).
Step 2: Verification by Solving Equation
Using quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$$x = \frac{-3 \pm \sqrt{-47}}{2(2)} = \frac{-3 \pm i\sqrt{47}}{4}$$
The roots contain $i = \sqrt{-1}$, confirming they are imaginary complex conjugates. Hence verified.
(iii) $16x^2 - 24x + 9 = 0$
Step 1: Determine Nature using Discriminant
$a = 16$, $b = -24$, $c = 9$
$$\text{Disc} = b^2 - 4ac = (-24)^2 - 4(16)(9) = 576 - 576 = 0$$
Since $\text{Disc} = 0$, the roots are real, rational, and equal.
Step 2: Verification by Solving Equation
$$(4x - 3)^2 = 0 \implies 4x - 3 = 0 \implies x = \frac{3}{4}$$
Thus, both roots are equal to $\frac{3}{4}$, which are real and rational. Hence verified.
(iv) $3x^2 + 7x - 13 = 0$
Step 1: Determine Nature using Discriminant
$a = 3$, $b = 7$, $c = -13$
$$\text{Disc} = b^2 - 4ac = (7)^2 - 4(3)(-13) = 49 + 156 = 205$$
Since $\text{Disc} = 205 > 0$ and $205$ is not a perfect square, the roots are real, irrational, and unequal.
Step 2: Verification by Solving Equation
$$x = \frac{-7 \pm \sqrt{205}}{2(3)} = \frac{-7 \pm \sqrt{205}}{6}$$
Since $\sqrt{205}$ cannot be simplified into an integer, the roots are irrational and distinct. Hence verified.
Question 3: For what value of $k$, the expression $k^2x^2 + 2(k+1)x + 4$ 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, $a = k^2$, $b = 2(k + 1)$, and $c = 4$.
Question 4: Find the value of $k$, if the roots of the following equations are equal.
(i) $(2k - 1)x^2 + 3kx + 3 = 0$
Solution:
For equal roots, $\text{Disc} = 0$.
$a = 2k - 1$, $b = 3k$, $c = 3$
$$\text{Disc} = b^2 - 4ac = (3k)^2 - 4(2k - 1)(3) = 0$$
$$9k^2 - 12(2k - 1) = 0$$
$$9k^2 - 24k + 12 = 0$$
Divide by $3$:
$$3k^2 - 8k + 4 = 0$$
Factorize:
$$3k^2 - 6k - 2k + 4 = 0$$
$$3k(k - 2) - 2(k - 2) = 0$$
$$(3k - 2)(k - 2) = 0$$
$$k = 2 \quad \text{or} \quad k = \frac{2}{3}$$
Answer: $k = 2, \frac{2}{3}$.
(ii) $x^2 + 2(k + 2)x + (3k + 4) = 0$
Solution:
$a = 1$, $b = 2(k + 2)$, $c = 3k + 4$
$$\text{Disc} = [2(k+2)]^2 - 4(1)(3k + 4) = 0$$
$$4(k^2 + 4k + 4) - 4(3k + 4) = 0$$
Divide by $4$:
$$(k^2 + 4k + 4) - (3k + 4) = 0$$
$$k^2 + k = 0$$
$$k(k + 1) = 0$$
$$k = 0 \quad \text{or} \quad k = -1$$
Answer: $k = 0, -1$.
Question 5: Show that the equation $x^2 + (mx + c)^2 = a^2$ has equal roots if $c^2 = a^2(1 + m^2)$.
Solution:
First, expand and write the equation in standard form $Ax^2 + Bx + C = 0$:
$$x^2 + (m^2x^2 + 2mcx + c^2) = a^2$$
$$(1 + m^2)x^2 + 2mcx + (c^2 - a^2) = 0$$
Here, $A = (1 + m^2)$, $B = 2mc$, and $C = (c^2 - a^2)$.
For equal roots, the discriminant must be zero ($\text{Disc} = 0$): $$\text{Disc} = B^2 - 4AC = 0$$ $$(2mc)^2 - 4(1 + m^2)(c^2 - a^2) = 0$$ $$4m^2c^2 - 4[c^2 - a^2 + m^2c^2 - m^2a^2] = 0$$ Divide throughout by $4$: $$m^2c^2 - (c^2 - a^2 + m^2c^2 - m^2a^2) = 0$$ $$m^2c^2 - c^2 + a^2 - m^2c^2 + m^2a^2 = 0$$ $$-c^2 + a^2 + m^2a^2 = 0$$ $$c^2 = a^2 + a^2m^2$$ $$c^2 = a^2(1 + m^2)$$ Hence proved.
Question 6: Find the condition that the roots of the equation $(c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0$ are equal.
Solution:
Comparing with $Ax^2 + Bx + C = 0$:
$A = c^2 - ab$, $B = -2(a^2 - bc)$, $C = b^2 - ac$.
For equal roots, $\text{Disc} = B^2 - 4AC = 0$: $$[-2(a^2 - bc)]^2 - 4(c^2 - ab)(b^2 - ac) = 0$$ $$4(a^2 - bc)^2 - 4(c^2 - ab)(b^2 - ac) = 0$$ Divide by $4$: $$(a^2 - bc)^2 - (c^2 - ab)(b^2 - ac) = 0$$ Expand both terms: $$(a^4 - 2a^2bc + b^2c^2) - (b^2c^2 - ac^3 - ab^3 + a^2bc) = 0$$ $$a^4 - 2a^2bc + b^2c^2 - b^2c^2 + ac^3 + ab^3 - a^2bc = 0$$ Combine like terms: $$a^4 + ab^3 + ac^3 - 3a^2bc = 0$$ Factor out $a$: $$a(a^3 + b^3 + c^3 - 3abc) = 0$$ Therefore, either: $$a = 0 \quad \text{or} \quad a^3 + b^3 + c^3 = 3abc$$ Answer: The required condition is $a = 0$ or $a^3 + b^3 + c^3 = 3abc$.
Q1: What is the discriminant of the quadratic equation $x^2 - 4x + 4 = 0$?
Q2: If the discriminant of a quadratic equation is negative ($\Delta < 0$), the roots are:
Q3: If $a, b, c$ are rational and $\Delta = 49$, the roots are:
Q4: For what value of $k$ does $x^2 + kx + 9 = 0$ have equal roots?
Q5: If the roots of $ax^2 + bx + c = 0$ are real and equal, then $b^2$ equals:
Q6: Irrational roots of a quadratic equation with rational coefficients always occur in:
Q7: The quadratic expression $4x^2 - 12x + 9$ is a:
Q8: If $\Delta = 20$ for a quadratic equation with rational coefficients, the roots are:
Q1: Why do students confuse rational and irrational roots when the discriminant is positive?
Ans: A common error in board examinations is assuming that any positive discriminant ($\Delta > 0$) implies rational roots. Students forget to check whether $\Delta$ is a perfect square. If $\Delta > 0$ and is a perfect square (e.g., $16, 25, 36$), the root term $\sqrt{\Delta}$ simplifies to an integer, making roots rational. If $\Delta > 0$ is NOT a perfect square (e.g., $8, 12, 17$), $\sqrt{\Delta}$ remains a surd, making roots irrational.
Q2: What mistake occurs when setting up conditions for perfect square expressions?
Ans: When asked to find unknown parameters (like $k$) that make a quadratic expression a perfect square (such as in Question 3), students often set the expression equal to zero directly or factorize incorrectly. The mathematically correct procedure is to calculate the discriminant of the expression and equate the discriminant to zero ($\text{Disc} = 0$), because a quadratic expression becomes $(px + q)^2$ if and only if its discriminant vanishes.
Q3: How do sign errors in $b^2 - 4ac$ lead to incorrect nature determinations?
Ans: When $b$ or $c$ is negative, students frequently make sign mistakes during squaring or multiplication. For example, in $2x^2 + 3x - 1 = 0$, students incorrectly calculate $b^2 - 4ac$ as $9 - 8 = 1$ instead of $9 - 4(2)(-1) = 9 + 8 = 17$. Always enclose negative numbers in parentheses when evaluating $b^2 - 4ac$.