Quadratic Equations Class 10 Important Questions with Solutions

Quadratic Equations is one of the most important chapters in Class 10 Maths because it combines formulas, factorisation, discriminant, application-based questions, and previous year board patterns in one chapter. It is also one of the most scoring chapters when students understand the standard methods properly and know how to identify what the question is really asking.

Many students feel comfortable with direct quadratic equations but get confused in questions based on discriminant, equal roots, reciprocal roots, word problems, and equations with radicals or fractions. That is why proper question practice matters more than only memorising formulas. Once students understand the logic behind root-based questions, the chapter becomes much easier.

At Deeksha Vedantu, we always encourage students to prepare Quadratic Equations through concept clarity first and question pattern recognition next. That approach makes board preparation much more effective.

Chapter Overview at a Glance

This quick table helps students revise the full chapter faster.

Quick Concept Table

Why Quadratic Equations Is Important in Class 10

This chapter is important because it appears in direct, application-based, and previous year question formats.

Why Students Must Prepare This Chapter Well

• it is a regular board-exam chapter
• it includes 2, 3, 4, and 5 mark questions
• it combines formulas and logical reasoning
• it helps improve algebraic accuracy
• it includes word problems and case-based formats

What Is a Quadratic Equation

A quadratic equation is an equation of degree 2.

Standard Form of a Quadratic Equation

ax² + bx + c = 0

Where:

• a, b, and c are real numbers
• a is not equal to zero

If a becomes zero, the equation will no longer remain quadratic.

Important Terms in Quadratic Equations

Students should know these terms clearly before solving board questions.

Terms Summary Table

Methods to Solve Quadratic Equations

There are three main methods used in Class 10.

Method Summary Table

Factorisation Method

This method is used when the quadratic equation can be broken into two simple factors.

Completing the Square Method

This method is more conceptual and helps derive the quadratic formula.

Quadratic Formula Method

This is one of the most important methods for board exams.

Core Formulas of Quadratic Equations

Students should memorise these formulas carefully.

Formula Sheet Table

Quadratic Formula

x = [-b ± √(b² – 4ac)] / 2a

Students must remember this formula accurately.

Discriminant

The expression inside the root sign in the quadratic formula is called the discriminant.

Its formula is:

D = b² – 4ac

This is very important because it tells us the nature of roots without fully solving the equation.

Nature of Roots Using Discriminant

This is one of the most frequently used concepts in previous year questions.

Case 1: D > 0

If D is greater than 0, the equation has two distinct real roots.

Case 2: D = 0

If D is equal to 0, the equation has two equal real roots.

Case 3: D < 0

If D is less than 0, the equation has no real roots.

Nature of Roots Table

Relation Between Roots and Coefficients

If α and β are the roots of the quadratic equation ax² + bx + c = 0, then the following relations hold.

Root-Coefficient Relation Table

These formulas are very useful in root-based questions.

Important Question 1: Find α + β + αβ

If α and β are the zeros of the polynomial 5x² – 6x + 1, find the value of α + β + αβ.

Given

For 5x² – 6x + 1:

• a = 5
• b = -6
• c = 1

Step 1: Use Root-Coefficient Relations

α + β = -b/a = -(-6)/5 = 6/5

αβ = c/a = 1/5

Step 2: Add the Values

α + β + αβ = 6/5 + 1/5 = 7/5

Answer

The required value is 7/5.

Important Question 2: Find the Discriminant and Comment on Nature of Roots

Find the discriminant of the quadratic equation 4x² – 5 = 0 and comment on the nature of roots.

Given

Equation: 4x² – 5 = 0

Step 1: Write the Equation in Standard Form

4x² + 0x – 5 = 0

So:

• a = 4
• b = 0
• c = -5

Step 2: Use the Discriminant Formula

D = b² – 4ac

D = 0² – 4 × 4 × (-5)

D = 80

Step 3: Comment on Nature of Roots

Since D > 0, the equation has two distinct real roots.

Answer

Discriminant = 80, and the equation has two distinct real roots.

Important Question 3: Reciprocal Roots Type Question

If one zero of the polynomial 6x² + 37x – (k – 2) is the reciprocal of the other, find the value of k.

Given

Polynomial: 6x² + 37x – (k – 2)

Step 1: Use the Reciprocal Roots Idea

If one root is α, the other is 1/α.

So their product is:

α × 1/α = 1

Step 2: Use Product of Roots Formula

For a quadratic equation:

αβ = c/a

Here:

c = -(k – 2) = -k + 2

a = 6

So:

(-k + 2)/6 = 1

Step 3: Solve

-k + 2 = 6

-k = 4

k = -4

Answer

The value of k is -4.

Important Question 4: Solve a Quadratic Equation with Fractions

Solve:

1/(x + 4) – 1/(x – 7) = 11/30

Given

Equation: 1/(x + 4) – 1/(x – 7) = 11/30

Step 1: Take LCM

LCM = (x + 4)(x – 7)

So:

[(x – 7) – (x + 4)] / [(x + 4)(x – 7)] = 11/30

Step 2: Simplify the Numerator

x – 7 – x – 4 = -11

So:

-11 / [(x + 4)(x – 7)] = 11/30

Step 3: Cross Multiply

-11 × 30 = 11 × (x + 4)(x – 7)

-30 = (x + 4)(x – 7)

Step 4: Expand

x² – 3x – 28 = -30

x² – 3x + 2 = 0

Step 5: Factorise

x² – 2x – x + 2 = 0

x(x – 2) – 1(x – 2) = 0

(x – 2)(x – 1) = 0

Answer

x = 2 or x = 1

Important Question 5: Form a Quadratic Polynomial from Reciprocal Roots

Find a quadratic polynomial whose zeros are the reciprocals of the zeros of the polynomial ax² + bx + c.

Given

Original polynomial: ax² + bx + c

Step 1: Assume the Original Roots

Let the original roots be m and n.

Then the required roots are:

• 1/m
• 1/n

Step 2: Find the Sum of New Roots

1/m + 1/n = (m + n)/mn

For the original polynomial:

m + n = -b/a

mn = c/a

So:

Sum of new roots = [(-b/a)] / (c/a) = -b/c

Step 3: Find the Product of New Roots

(1/m)(1/n) = 1/mn

Since mn = c/a,

Product of new roots = a/c

Step 4: Form the Polynomial

x² – (sum of roots)x + product of roots

x² – (-b/c)x + a/c

x² + (b/c)x + a/c

Multiply through by c:

cx² + bx + a

Answer

The required quadratic polynomial is cx² + bx + a.

Important Question 6: Solve a Quadratic Equation with Radicals

Solve:

√3 x² – 2√2 x – 2√3 = 0

Given

Equation: √3 x² – 2√2 x – 2√3 = 0

Step 1: Identify the Pattern

This type of equation is usually easier through factorisation after careful splitting.

Step 2: Rewrite the Middle Term

A suitable split is:

√3 x² – 3√2 x + √2 x – 2√3 = 0

Step 3: Group the Terms

(√3 x² – 3√2 x) + (√2 x – 2√3) = 0

This can be arranged into factor form:

(x – √2)(√3 x + 2) = 0

Step 4: Find the Roots

x – √2 = 0 gives x = √2

√3 x + 2 = 0 gives x = -2/√3

Answer

The roots are √2 and -2/√3.

Important Question 7: Word Problem Based on Speed of a Plane

A plane left 30 minutes later than its scheduled time and to reach the destination 1500 km away on time, it had to increase its speed by 100 km per hour from its usual speed. Find its usual speed.

Given

• Distance = 1500 km
• Delay = 30 minutes = 1/2 hour
• Increased speed = usual speed + 100

Step 1: Let the Usual Speed Be x km/h

Then the usual time taken is:

1500/x

Step 2: Write the New Time

New speed = x + 100

New time taken = 1500/(x + 100)

Step 3: Use the Time Difference

1500/x – 1500/(x + 100) = 1/2

Step 4: Simplify

Take LCM and cross multiply:

1500(x + 100) – 1500x = x(x + 100)/2

150000 = x(x + 100)/2

300000 = x² + 100x

x² + 100x – 300000 = 0

Step 5: Factorise

x² + 600x – 500x – 300000 = 0

x(x + 600) – 500(x + 600) = 0

(x + 600)(x – 500) = 0

Step 6: Accept the Positive Value

x = 500 or x = -600

Speed cannot be negative, so x = 500

Answer

The usual speed of the plane is 500 km/h.

Important Question 8: Equal Roots Condition

Find the value of k for which the quadratic equation (k + 1)x² – 6(k + 1)x + 3(k + 9) = 0 has equal roots.

Given

Equation: (k + 1)x² – 6(k + 1)x + 3(k + 9) = 0

Step 1: Use the Equal Roots Condition

For equal roots:

D = 0

Step 2: Identify Coefficients

• a = k + 1
• b = -6(k + 1)
• c = 3(k + 9)

Step 3: Apply the Discriminant Formula

b² – 4ac = 0

[-6(k + 1)]² – 4(k + 1) × 3(k + 9) = 0

36(k + 1)² – 12(k + 1)(k + 9) = 0

Step 4: Simplify

Take 12(k + 1) common:

12(k + 1)[3(k + 1) – (k + 9)] = 0

12(k + 1)(2k – 6) = 0

(k + 1)(k – 3) = 0

So:

k = -1 or k = 3

Step 5: Check Validity

If k = -1, then a = k + 1 = 0, so the equation stops being quadratic.

So the valid value is:

k = 3

Answer

The valid value of k is 3.

Important Question 9: Find p and k from Root Information

If -5 is a root of the equation 2x² + px – 5 = 0 and the equation px² + px + k = 0 has equal roots, find p and k.

Given

• First equation: 2x² + px – 5 = 0
• Root of first equation: x = -5
• Second equation: px² + px + k = 0 has equal roots

Step 1: Use the First Equation

Put x = -5 in 2x² + px – 5 = 0

2(25) – 5p – 5 = 0

50 – 5p – 5 = 0

45 – 5p = 0

p = 9

Step 2: Use the Second Equation

Now the equation becomes:

9x² + 9x + k = 0

Since it has equal roots, D = 0

Step 3: Apply Discriminant

9² – 4 × 9 × k = 0

81 – 36k = 0

36k = 81

k = 9/4

Answer

p = 9 and k = 9/4

Board Question Patterns Students Must Practise

The chapter often repeats similar categories of questions.

Board Pattern Summary Table

Common Mistakes Students Make in Quadratic Equations

Common Mistakes Table

Quick Formula Sheet for Quadratic Equations

This section is useful for last-minute revision.

Quick Formula Table

Nature of Roots Quick Table

Best Study Strategy for Quadratic Equations

Students can score much better in this chapter when revision is done methodically.

Step-by-Step Revision Table

Practice Questions for Students

Important Practice Questions

• Solve x² – 7x + 12 = 0.
• Find the discriminant of 3x² + 5x + 2 = 0.
• Find the value of k if the equation x² + kx + 9 = 0 has equal roots.
• Form a quadratic polynomial whose roots are 3 and 5.
• If α and β are roots of 2x² – 9x + 4 = 0, find α + β and αβ.
• Solve a word problem based on speed and delay using a quadratic equation.

FAQs

Q1. What is a quadratic equation in Class 10 Maths?

A quadratic equation is an equation of degree 2 written in the form ax² + bx + c = 0, where a ≠ 0.

Q2. What is the formula for solving quadratic equations?

The quadratic formula is x = [-b ± √(b² – 4ac)] / 2a.

Q3. What is the discriminant?

The discriminant is b² – 4ac. It tells us the nature of roots of the quadratic equation.

Q4. When does a quadratic equation have equal roots?

A quadratic equation has equal roots when the discriminant is equal to zero.

Q5. What are the relations between roots and coefficients?

If α and β are the roots of ax² + bx + c = 0, then α + β = -b/a and αβ = c/a.

Q6. Why do students find quadratic word problems difficult?

Students often find them difficult because they rush while reading the question and do not define the variable clearly at the start.

Q7. Can a quadratic equation have no real roots?

Yes. If the discriminant is less than zero, the quadratic equation has no real roots.

Q8. How can I score well in Quadratic Equations in boards?

You can score well by memorising formulas, practising standard question types, revising sign rules carefully, and solving previous year style questions regularly.

Conclusion

Quadratic Equations is one of the most important and scoring chapters in Class 10 Maths when students prepare it through formulas, root concepts, discriminant logic, and repeated board-style practice. The chapter may look lengthy because of the variety of question types, but once the pattern becomes familiar, most questions become manageable.

The smartest way to master this chapter is to revise the formulas well, solve different kinds of important questions, and practise application-based problems without panic. At Deeksha Vedantu, we always encourage students to focus on understanding the method behind each question, because that is what builds long-term confidence in board preparation.