Quadratic Equations questions with answers and concepts

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107 practice questionsPage updated 11 Nov 2026Review metadata pendingPractice questions

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A quadratic equation is a second-degree polynomial equation of the single variable x with standard form ax² + $b x + c = 0$ , where a ≠ 0 and a, b, c are real numbers. By the Fundamental Theorem of Algebra, every quadratic equation has exactly two roots (which may be real and distinct, real and coincident/equal, or complex conjugates). Solving quadratic equations and analyzing their roots is a fundamental building block for algebra, coordinate geometry, calculus, and competitive aptitude examinations including CAT, GATE, and banking exams.

Key formulas and rules

11 rules

Standard Form: $a x^{2} + b x + c = 0$ where $a \neq = 0$

Quadratic Formula (Shreedharacharya's Rule): $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

Discriminant: $Δ = D = b^{2} - 4 a c$

Nature of Roots: If $Δ > 0$ , roots are real and distinct; if $Δ = 0$ , roots are real and equal ( $x = - \frac{b}{2 a}$ ); if $Δ < 0$ , roots are non-real complex conjugates

Rational Roots: If $a, b, c \in Q$ and $Δ$ is a perfect square, roots are rational; if $Δ$ is not a perfect square, roots occur in conjugate irrational pairs ( $p \pm q$ )

Vieta's Formulas (Sum and Product): If $α, β$ are roots, then $α + β = - \frac{b}{a}$ and $α β = \frac{c}{a}$

Forming Quadratic Equation: $x^{2} - (α + β) x + α β = 0$ , or $x^{2} - (Sum of roots) x + (Product of roots) = 0$

Symmetric Root Functions: $α^{2} + β^{2} = (α + β)^{2} - 2 α β$ , and $(α - β)^{2} = (α + β)^{2} - 4 α β = \frac{Δ}{a ^{2}}$

Sum of Reciprocals: $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = - \frac{b}{c}$ (for $c \neq = 0$ )

Vertex / Extrema of $f (x) = a x^{2} + b x + c$ : Vertex at $x = - \frac{b}{2 a}$ . If $a > 0$ , minimum is $\frac{4 a c - b ^{2}}{4 a} = - \frac{Δ}{4 a}$ ; if $a < 0$ , maximum is $- \frac{Δ}{4 a}$

Condition for a Common Root between $a_{1} x^{2} + b_{1} x + c_{1} = 0$ and $a_{2} x^{2} + b_{2} x + c_{2} = 0$ : $(c_{1} a_{2} - c_{2} a_{1})^{2} = (a_{1} b_{2} - a_{2} b_{1}) (b_{1} c_{2} - b_{2} c_{1})$

Key concepts

Solving by Factorization (Splitting the Middle Term)

Nature of Roots via Discriminant

Vieta's Formulas and Symmetric Expressions

Sign of Quadratic Expression

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Worked examples

Example 1

Find the roots of the quadratic equation $2 x^{2} - 7 x + 3 = 0$ .

Solution

Step 1: Identify coefficients: $a = 2, b = - 7, c = 3$ .
Step 2: Find two numbers whose sum is $- 7$ and product is $a \times c = 6$ . The numbers are $- 6$ and $- 1$ .
Step 3: Split the middle term:
$2 x^{2} - 6 x - x + 3 = 0$
$2 x (x - 3) - 1 (x - 3) = 0$
$(2 x - 1) (x - 3) = 0$
Step 4: Set each factor to 0:
$2 x - 1 = 0 ⟹ x = 1/2$
$x - 3 = 0 ⟹ x = 3$

Answer: The roots are $x = 1/2$ and $x = 3$ .

Example 2

For what values of $k$ does the equation $x^{2} + 2 (k - 1) x + (k + 5) = 0$ have real and equal roots?

Solution

Step 1: For real and equal roots, the discriminant $Δ = b^{2} - 4 a c$ must equal 0.
Step 2: Here $a = 1$ , $b = 2 (k - 1)$ , $c = k + 5$ .
$Δ = [2 (k - 1)]^{2} - 4 (1) (k + 5) = 0$
$4 (k^{2} - 2 k + 1) - 4 (k + 5) = 0$
Divide by 4:
$k^{2} - 2 k + 1 - k - 5 = 0$
$k^{2} - 3 k - 4 = 0$
Step 3: Factor the quadratic in $k$ :
$(k - 4) (k + 1) = 0$
$k = 4$ or $k = - 1$ .

Answer: $k = 4$ or $k = - 1$ .

Example 3

If $α$ and $β$ are the roots of $x^{2} - 5 x + 6 = 0$ , find the value of $α^{2} + β^{2}$ and $\frac{1}{α} + \frac{1}{β}$ .

Solution

Step 1: By Vieta's formulas, $α + β = - (- 5) /1 = 5$ and $α β = 6/1 = 6$ .
Step 2: Calculate $α^{2} + β^{2}$ :
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = 5^{2} - 2 (6) = 25 - 12 = 13$ .
Step 3: Calculate $\frac{1}{α} + \frac{1}{β}$ :
$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{5}{6}$ .

Answer: $α^{2} + β^{2} = 13$ and $\frac{1}{α} + \frac{1}{β} = \frac{5}{6}$ .

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

IBPS PO 2020 Shift 1

In the question, two equations (I) and (II) are given. You have to solve both equations to establish the correct relation between x and y.

I. x² - $3 x - 88 = 0$
II. y² + $8 y - 48 = 0$

x > y
x >= y
x < y
x <= y
$x = y$ or relation cannot be established

Answer: E. x = y or relation cannot be established

Explanation

Solving Equation I: x² - $3 x - 88 = 0$ gives $(x - 11) (x + 8) = 0$ , so $x = 11$ or x = -8.
Solving Equation II: y² + $8 y - 48 = 0$ gives $(y + 12) (y - 4) = 0$ , so $y = 4$ or y = -12.
Comparing the roots: When $x = 11$ , 11 > 4 and 11 > -12. When x = -8, -8 < 4 and -8 > -12. Because x is greater than y for some roots and smaller for others, no definite relation can be established.

Sources and review notes

This question retains its previous-year paper provenance.

IBPS PO 2020 · IBPS PO 2020 Prelims · Question 1Question paper

Review status: verified · Reviewed 2026-08-13 · structure and answer-key checks, worked-answer review, human editorial review, duplicate screening

Question 2

IBPS PO 2020 Shift 1

In the question, two equations (I) and (II) are given. You have to solve both equations to establish the correct relation between x and y.

I. 5x² + $29 x + 20 = 0$
II. 25y² + $25 y + 6 = 0$

x > y
x >= y
x < y
x <= y
$x = y$ or relation cannot be established

Answer: C. x < y

Explanation

Solving Equation I: 5x² + $25 x + 4 x + 20 = 0$ gives $(5 x + 4) (x + 5) = 0$ , so x = -5 or x = -0.8.
Solving Equation II: 25y² + $15 y + 10 y + 6 = 0$ gives $(5 y + 3) (5 y + 2) = 0$ , so y = -0.6 or y = -0.4.
Comparing the roots: -5 < -0.6 and -5 < -0.4. Also -0.8 < -0.6 and -0.8 < -0.4. Across all comparisons, x < y holds strictly.

Sources and review notes

This question retains its previous-year paper provenance.

IBPS PO 2020 · IBPS PO 2020 Prelims · Question 2Question paper

Review status: verified · Reviewed 2026-08-13 · structure and answer-key checks, worked-answer review, human editorial review, duplicate screening

Question 3

IBPS PO 2020 Shift 1

In the question, two equations (I) and (II) are given. You have to solve both equations to establish the correct relation between x and y.

I. 6x² + $13 x + 5 = 0$
II. 9y² + $22 y + 8 = 0$

x > y
x >= y
x < y
x <= y
$x = y$ or relation cannot be established

Answer: E. x = y or relation cannot be established

Explanation

Solving Equation I: 6x² + $10 x + 3 x + 5 = 0$ gives $(2 x + 1) (3 x + 5) = 0$ , which yields x = -0.5 and x = -1.67.
Solving Equation II: 9y² + $18 y + 4 y + 8 = 0$ gives $(9 y + 4) (y + 2) = 0$ , which yields y = -0.44 and y = -2.
Comparing roots: For x = -0.5 and y = -0.44, x < y; but for x = -0.5 and y = -2, x > y. Therefore, no relation can be established between x and y.

Sources and review notes

This question retains its previous-year paper provenance.

IBPS PO 2020 · IBPS PO 2020 Prelims · Question 3Question paper

Review status: verified · Reviewed 2026-08-13 · structure and answer-key checks, worked-answer review, human editorial review, duplicate screening

Common mistakes and useful habits

Always check if $a + b + c = 0$: if so, $x = 1$ is immediately one root and $x = c/a$ is the second root.
If $a - b + c = 0$, then $x = -1$ is one root and $x = -c/a$ is the second root.
If the roots are reciprocal of each other ($\alpha = 1/\beta$), then $\alpha\beta = 1 \implies c/a = 1 \implies c = a$.
If roots are equal in magnitude but opposite in sign ($\alpha = -\beta$), then $\alpha + \beta = 0 \implies -b/a = 0 \implies b = 0$.
When taking square roots in $x^2 = k$, remember both $+ \sqrt{k}$ and $- \sqrt{k}$.

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Worked examples

Example 1

Find the roots of the quadratic equation $2 x^{2} - 7 x + 3 = 0$ .

Solution

Answer: The roots are $x = 1/2$ and $x = 3$ .

Example 2

For what values of $k$ does the equation $x^{2} + 2 (k - 1) x + (k + 5) = 0$ have real and equal roots?

Solution

Answer: $k = 4$ or $k = - 1$ .

Example 3

If $α$ and $β$ are the roots of $x^{2} - 5 x + 6 = 0$ , find the value of $α^{2} + β^{2}$ and $\frac{1}{α} + \frac{1}{β}$ .

Solution

Answer: $α^{2} + β^{2} = 13$ and $\frac{1}{α} + \frac{1}{β} = \frac{5}{6}$ .

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

IBPS PO 2020 Shift 1

In the question, two equations (I) and (II) are given. You have to solve both equations to establish the correct relation between x and y.

I. x² - $3 x - 88 = 0$
II. y² + $8 y - 48 = 0$

x > y
x >= y
x < y
x <= y
$x = y$ or relation cannot be established

Answer: E. x = y or relation cannot be established

Explanation

Sources and review notes

This question retains its previous-year paper provenance.

IBPS PO 2020 · IBPS PO 2020 Prelims · Question 1Question paper

Review status: verified · Reviewed 2026-08-13 · structure and answer-key checks, worked-answer review, human editorial review, duplicate screening

Question 2

IBPS PO 2020 Shift 1

In the question, two equations (I) and (II) are given. You have to solve both equations to establish the correct relation between x and y.

I. 5x² + $29 x + 20 = 0$
II. 25y² + $25 y + 6 = 0$

x > y
x >= y
x < y
x <= y
$x = y$ or relation cannot be established

Answer: C. x < y

Explanation

Sources and review notes

This question retains its previous-year paper provenance.

IBPS PO 2020 · IBPS PO 2020 Prelims · Question 2Question paper

Review status: verified · Reviewed 2026-08-13 · structure and answer-key checks, worked-answer review, human editorial review, duplicate screening

Question 3

IBPS PO 2020 Shift 1

In the question, two equations (I) and (II) are given. You have to solve both equations to establish the correct relation between x and y.

I. 6x² + $13 x + 5 = 0$
II. 9y² + $22 y + 8 = 0$

x > y
x >= y
x < y
x <= y
$x = y$ or relation cannot be established

Answer: E. x = y or relation cannot be established

Explanation

Sources and review notes

This question retains its previous-year paper provenance.

IBPS PO 2020 · IBPS PO 2020 Prelims · Question 3Question paper

Review status: verified · Reviewed 2026-08-13 · structure and answer-key checks, worked-answer review, human editorial review, duplicate screening

Common mistakes and useful habits

Always check if $a + b + c = 0$: if so, $x = 1$ is immediately one root and $x = c/a$ is the second root.
If $a - b + c = 0$, then $x = -1$ is one root and $x = -c/a$ is the second root.
If the roots are reciprocal of each other ($\alpha = 1/\beta$), then $\alpha\beta = 1 \implies c/a = 1 \implies c = a$.
If roots are equal in magnitude but opposite in sign ($\alpha = -\beta$), then $\alpha + \beta = 0 \implies -b/a = 0 \implies b = 0$.
When taking square roots in $x^2 = k$, remember both $+ \sqrt{k}$ and $- \sqrt{k}$.

Key formulas and rules

Key concepts

Solving by Factorization (Splitting the Middle Term)

Nature of Roots via Discriminant

Vieta's Formulas and Symmetric Expressions

Sign of Quadratic Expression

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

Keep building the same skill family.

Key formulas and rules

Key concepts

Solving by Factorization (Splitting the Middle Term)

Nature of Roots via Discriminant

Vieta's Formulas and Symmetric Expressions

Sign of Quadratic Expression

Worked examples

Example 1

Solution

Example 2

Solution

Example 3

Solution

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

Keep building the same skill family.