Quadratic Equation Examples for Beginners with Solutions

Why Examples Are the Best Way to Learn Quadratics

Reading rules about quadratic equations can feel dry. Solving examples is where the ideas become clear. In this article you will meet six short examples, each showing a different type of quadratic equation, followed by three longer word problems with Indian numbers.

Every standard quadratic equation looks like ax^2 + bx + c = 0. Keep that form in mind as you read. When you finish an example by hand, you can check it with a free quadratic equation calculator.

Six Quick Examples, One for Each Type

Type 1: No middle term, x^2 – 49 = 0

Move 49 across: x^2 = 49. Take the square root of both sides. Remember both signs. So x = 7 or x = -7.

Type 2: No constant term, x^2 – 3x = 0

Take x common: x(x – 3) = 0. Either x = 0 or x – 3 = 0. So the roots are 0 and 3. Never divide both sides by x here, or you will lose the root x = 0.

Type 3: Easy factorisation, x^2 + 7x + 12 = 0

We need two numbers with product 12 and sum 7. They are 3 and 4. So (x + 3)(x + 4) = 0, and x = -3 or x = -4.

Type 4: When a is not 1, 2x^2 + 5x – 3 = 0

Use the formula with a = 2, b = 5, c = -3. D = 25 + 24 = 49, and the square root is 7. Then x = (-5 + 7) / 4 = 0.5 and x = (-5 – 7) / 4 = -3. Check the sum: 0.5 + (-3) = -2.5, and -b/a = -5/2 = -2.5.

Type 5: Perfect square, x^2 + 10x + 25 = 0

This is (x + 5)^2 = 0, so x = -5 only. The discriminant is 100 – 100 = 0, so the two roots are equal.

Type 6: No real roots, x^2 + x + 1 = 0

D = 1 – 4 = -3. Since D is negative, there are no real roots. This is a valid answer, not a mistake.

Equation	Method used	Roots
x^2 – 49 = 0	Square root	7 and -7
x^2 – 3x = 0	Take x common	0 and 3
x^2 + 7x + 12 = 0	Factorisation	-3 and -4
2x^2 + 5x – 3 = 0	Formula	0.5 and -3
x^2 + 10x + 25 = 0	Perfect square	-5 (repeated)
x^2 + x + 1 = 0	Discriminant	No real roots

How to Turn a Word Problem into an Equation

1. Read the problem twice and decide what you need to find.
2. Let that quantity be x and name its unit.
3. Write other quantities in terms of x.
4. Use the given fact, such as an area, a product or a total, to form an equation.
5. Simplify to ax^2 + bx + c = 0 and solve.
6. Reject any root that does not fit the story, then check the answer.

How to Spot the Type Quickly

Before you start any calculation, spend ten seconds looking at the equation. This habit saves more time than any trick.

• If there is no x term, as in x^2 – 49 = 0, move the number across and take square roots.
• If there is no constant, as in x^2 – 3x = 0, take x common and set each factor to zero.
• If a = 1 and the numbers are small, look for two numbers with the right product and sum.
• If a is not 1 or the numbers are large, go straight to the formula.
• If the whole equation can be divided by a common number, do that first to make the numbers smaller.

How to Check Any Answer in Ten Seconds

Put each root back into the original equation. For x^2 + 7x + 12 = 0 and x = -3, the left side is 9 – 21 + 12 = 0, which is correct. For x = -4 it is 16 – 28 + 12 = 0, also correct.

You can also use the roots’ sum and product. Here the roots add to -7, and -b/a is -7. They multiply to 12, and c/a is 12. If either test fails, go back and look for a sign error.

Three Worked Examples with Indian Numbers

Example 1: Price of a pen

A trader buys some pens for Rs. 480. If each pen had cost Rs. 4 less, he would have got 20 more pens for the same money. Find the price of one pen.

Let the price be Rs. p. The number of pens is 480/p. At the lower price, the number is 480/(p – 4). The difference is 20: 480/(p – 4) – 480/p = 20.

Combine: 480p – 480(p – 4) = 20p(p – 4). The left side is 1,920, so 1,920 = 20p^2 – 80p. Divide by 20: 96 = p^2 – 4p, so p^2 – 4p – 96 = 0.

D = 16 + 384 = 400, and its square root is 20. Then p = (4 + 20) / 2 = 12 or p = (4 – 20) / 2 = -8. A price cannot be negative, so each pen costs Rs. 12. Check: 480 / 12 = 40 pens, and 480 / 8 = 60 pens. The difference is 20.

Example 2: A ladder against a wall

A 13 m ladder leans against a wall. The top of the ladder reaches 7 m higher than the distance of its foot from the wall. Find that distance.

Let the distance be x. The height is x + 7. By the Pythagoras theorem, x^2 + (x + 7)^2 = 13^2. Expand: x^2 + x^2 + 14x + 49 = 169, so 2x^2 + 14x – 120 = 0. Divide by 2: x^2 + 7x – 60 = 0.

D = 49 + 240 = 289, and its square root is 17. Then x = (-7 + 17) / 2 = 5 or x = (-7 – 17) / 2 = -12. Distance cannot be negative, so x = 5 m and the height is 12 m. This is the famous 5-12-13 triangle: 25 + 144 = 169.

Example 3: Rahul’s age

The product of Rahul’s age 3 years ago and his age 5 years from now is 105. Find his present age.

Let his present age be x years. Then (x – 3)(x + 5) = 105. Expand: x^2 + 2x – 15 = 105, so x^2 + 2x – 120 = 0.

D = 4 + 480 = 484, and its square root is 22. Then x = (-2 + 22) / 2 = 10 or x = (-2 – 22) / 2 = -12. Age cannot be negative, so Rahul is 10 years old. Check: 7 x 15 = 105.

Key takeaway: In every word problem the steps are the same: define x, build the equation, solve it and reject roots that break the story.

Practice Set with Answers

Equation	Answer
x^2 – 9x + 20 = 0	x = 4 or x = 5
x^2 + 2x – 24 = 0	x = 4 or x = -6
3x^2 – 12 = 0	x = 2 or x = -2
x^2 – 6x + 10 = 0	No real roots (D = -4)

Benefits of Practising Examples

• Pattern recognition: After a few examples you can tell at a glance which method to use.
• Exam speed: Board questions repeat the same patterns, so practice saves minutes.
• Confidence: Seeing the check work out builds trust in your own answers.
• Understanding word problems: Price, age and geometry questions become routine.
• Strong base: Easy examples prepare you for harder JEE-level algebra.

Challenges and Limitations

Beginner examples are tidy, with neat whole-number roots. Real questions can have fractions, surds or large numbers, and these take more care. Word problems are especially tough because the equation is not given to you.

Copying solved examples without understanding them is another trap. If you only memorise steps, a slightly changed question may stop you. Spend time on why each step is allowed. For decimal answers, the decimal calculator can help you check your arithmetic.

Common Mistakes to Avoid

• Losing the negative root: From x^2 = 49, both 7 and -7 are answers.
• Dividing by x: In x^2 – 3x = 0, dividing by x loses the root x = 0.
• Wrong factor signs: For x^2 + 7x + 12, the factors are (x + 3) and (x + 4), so the roots are negative.
• Not moving everything to one side: The right side must be zero before factorising.
• Arithmetic errors in D: Write b^2 and 4ac separately, then subtract.
• Forgetting units in word problems: End your answer with Rs., m or years.

Best Practices

• Solve at least one example of each type until the methods feel natural.
• Write the standard form and list a, b, c before solving.
• Use the sum and product of roots to check: -b/a and c/a.
• Draw a small diagram for geometry problems, like the ladder.
• Reject impossible roots and say why in one short line.
• Compare your final answers with an online calculator, and study any difference.

Conclusion

Quadratic equations become friendly once you have seen the main types. Square roots handle equations with no middle term, common factors handle those with no constant, factorisation handles small whole numbers and the formula handles everything else.

Work through the six quick examples and three word problems again on your own, then try the practice set. A little daily practice is all it takes to make quadratics a strength in Class 10 and beyond.

Frequently Asked Questions

What is a simple example of a quadratic equation?

A simple example is x^2 – 5x + 6 = 0. It factorises as (x – 2)(x – 3) = 0, so the roots are x = 2 and x = 3. You can check by putting each value back in the equation.

How do I know which method to use?

If the numbers are small, try factorisation first. If there is no middle term, take square roots. For anything else, or when you are unsure, use the quadratic formula because it always works.

Can the answer to a quadratic equation be zero?

Yes. In x^2 – 3x = 0, the roots are 0 and 3. This happens whenever the constant term c is zero.

What should I do if both roots look correct in a word problem?

Check each one against the story. If a root gives a negative length, age or price, reject it. If both roots make sense, then both are valid answers.

How many practice problems should a beginner solve?

Aim for five to ten problems a day for a week, mixing the types. After solving each by hand, verify with a calculator and note any mistakes.