Quadratic Equations Notes

4

Quadratic Equations

Detailed Theory • Easy English • Board Exam Focus

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Chapter Roadmap

1. Basic Idea and Standard Form 2. Roots of a Quadratic Equation 3. Solving by Factorisation 4. Quadratic Formula 5. Discriminant and Nature of Roots 6. Relationship Between Roots and Coefficients 7. Word Problems 8. Board-Style Questions 9. Board Exam Strategy 10. Final 96% Target Revision

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1. Basic Idea and Standard Form

What is a Quadratic Equation?

A quadratic equation in one variable is an equation of the form

\[\boxed{ax^2+bx+c=0,\quad a\neq0}\]

Here \(a,b,c\) are real numbers and \(a\neq0\). The highest power of the variable is \(2\).

Solved Example 1 — Identify \(a,b,c\)

Question: Identify \(a,b,c\) in \(3x^2-7x+5=0\).

1. Compare the equation with \(ax^2+bx+c=0\).
2. The coefficient of \(x^2\) is \(3\), so \(a=3\).
3. The coefficient of \(x\) is \(-7\), so \(b=-7\).
4. The constant term is \(5\), so \(c=5\).
5. Answer: \(\boxed{a=3,\ b=-7,\ c=5}\).

Solved Example 2 — Check whether it is quadratic

Question: Is \(5x^2+2x-8=0\) a quadratic equation?

1. Look at the highest power of \(x\).
2. The highest power is \(2\).
3. The coefficient of \(x^2\) is \(5\), which is not zero.
4. Therefore, it has the standard quadratic form.
5. Answer: Yes, it is a \(\boxed{\text{quadratic equation}}\).

Solved Example 3 — Convert to standard form

Question: Convert \(x^2+6=5x\) into standard form.

1. Start with \(x^2+6=5x\).
2. Bring \(5x\) to the left side.
3. \(x^2-5x+6=0\).
4. Now the equation is in the form \(ax^2+bx+c=0\).
5. Answer: \(\boxed{x^2-5x+6=0}\).

2. Roots of a Quadratic Equation

Meaning of Root

A number \(\alpha\) is called a root or solution of a quadratic equation if it makes the equation true.

\[\boxed{a\alpha^2+b\alpha+c=0}\]

Solved Example 1 — Check a root

Question: Check whether \(2\) is a root of \(x^2-5x+6=0\).

1. Substitute \(x=2\).
2. \(2^2-5(2)+6\).
3. \(=4-10+6\).
4. \(=0\).
5. Since the result is \(0\), \(2\) is a root.
6. Answer: \(\boxed{2}\) is a root.

Solved Example 2 — Check another number

Question: Check whether \(1\) is a root of \(2x^2-3x+5=0\).

1. Substitute \(x=1\).
2. \(2(1)^2-3(1)+5\).
3. \(=2-3+5\).
4. \(=4\neq0\).
5. Therefore, \(1\) does not satisfy the equation.
6. Answer: \(\boxed{1}\) is not a root.

Solved Example 3 — Verify a root

Question: Verify that \(3\) is a root of \(x^2-6x+9=0\).

1. Put \(x=3\).
2. \(3^2-6(3)+9\).
3. \(=9-18+9\).
4. \(=0\).
5. Hence the equation is satisfied.
6. Answer: \(\boxed{3}\) is a root.

3. Solving a Quadratic Equation by Factorisation

Basic Method

To solve by factorisation, first write the equation in standard form, factorise it, and then use the zero-product property.

\[\boxed{uv=0\Rightarrow u=0\ \text{or}\ v=0}\]

Solved Example 1 — Simple factorisation

Question: Solve \(x^2-5x+6=0\).

1. Find two numbers whose product is \(6\) and sum is \(-5\).
2. The numbers are \(-2\) and \(-3\).
3. Therefore, \(x^2-5x+6=(x-2)(x-3)\).
4. So \((x-2)(x-3)=0\).
5. Therefore, \(x-2=0\) or \(x-3=0\).
6. Thus \(x=2\) or \(x=3\).
7. Answer: \(\boxed{x=2,3}\).

Solved Example 2 — Negative roots

Question: Solve \(x^2+x-6=0\).

1. We need two numbers whose product is \(-6\) and sum is \(1\).
2. The numbers are \(3\) and \(-2\).
3. So \(x^2+x-6=(x+3)(x-2)\).
4. Hence \((x+3)(x-2)=0\).
5. Therefore, \(x+3=0\) or \(x-2=0\).
6. So \(x=-3\) or \(x=2\).
7. Answer: \(\boxed{x=-3,2}\).

Solved Example 3 — Coefficient greater than 1

Question: Solve \(2x^2-7x+3=0\).

1. Multiply \(a\) and \(c\): \(2\times3=6\).
2. Find two numbers whose product is \(6\) and sum is \(-7\): \(-6,-1\).
3. Split the middle term: \(2x^2-6x-x+3=0\).
4. Group the terms: \(2x(x-3)-1(x-3)=0\).
5. Take common factor: \((2x-1)(x-3)=0\).
6. Therefore, \(2x-1=0\) or \(x-3=0\).
7. So \(x=\frac12\) or \(x=3\).
8. Answer: \(\boxed{x=\frac12,3}\).

4. Solving by Quadratic Formula

The Formula

For \(ax^2+bx+c=0\), the roots are found using

\[\boxed{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\]

The symbol \(\pm\) gives two possible values, one with \(+\) and one with \(-\).

Solved Example 1 — Two real roots

Question: Solve \(x^2-5x+6=0\) using the quadratic formula.

1. Compare with \(ax^2+bx+c=0\): \(a=1,\ b=-5,\ c=6\).
2. Use \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\).
3. \(x=\frac{5\pm\sqrt{25-24}}{2}\).
4. \(x=\frac{5\pm1}{2}\).
5. First root: \(x=\frac{5+1}{2}=3\).
6. Second root: \(x=\frac{5-1}{2}=2\).
7. Answer: \(\boxed{x=2,3}\).

Solved Example 2 — Irrational roots

Question: Solve \(x^2-4x-1=0\).

1. Here \(a=1,\ b=-4,\ c=-1\).
2. Substitute in the formula.
3. \(x=\frac{4\pm\sqrt{16+4}}{2}\).
4. \(x=\frac{4\pm\sqrt{20}}{2}\).
5. \(\sqrt{20}=2\sqrt5\).
6. Therefore \(x=\frac{4\pm2\sqrt5}{2}\).
7. So \(x=2\pm\sqrt5\).
8. Answer: \(\boxed{x=2+\sqrt5,\ 2-\sqrt5}\).

Solved Example 3 — Repeated root

Question: Solve \(x^2-6x+9=0\) using the formula.

1. Here \(a=1,\ b=-6,\ c=9\).
2. \(x=\frac{6\pm\sqrt{36-36}}{2}\).
3. \(x=\frac{6\pm0}{2}\).
4. Both signs give the same value.
5. \(x=3\).
6. Answer: The repeated root is \(\boxed{3}\).

5. Discriminant and Nature of Roots

Discriminant

The expression under the square root in the quadratic formula is called the discriminant.

\[\boxed{D=b^2-4ac}\]

Nature of roots:

• If \(D>0\), there are two distinct real roots.
• If \(D=0\), there are two equal real roots.
• If \(D<0\), there are no real roots.

Solved Example 1 — \(D>0\)

Question: Find the nature of roots of \(x^2-5x+6=0\).

1. Here \(a=1,\ b=-5,\ c=6\).
2. Calculate \(D=b^2-4ac\).
3. \(D=(-5)^2-4(1)(6)\).
4. \(D=25-24=1\).
5. Since \(D>0\), the roots are real and distinct.
6. Answer: \(\boxed{\text{Two distinct real roots}}\).

Solved Example 2 — \(D=0\)

Question: Find the nature of roots of \(x^2-6x+9=0\).

1. Here \(a=1,\ b=-6,\ c=9\).
2. \(D=(-6)^2-4(1)(9)\).
3. \(D=36-36=0\).
4. Since \(D=0\), both roots are equal and real.
5. Answer: \(\boxed{\text{Two equal real roots}}\).

Solved Example 3 — \(D<0\)

Question: Find the nature of roots of \(x^2+2x+5=0\).

1. Here \(a=1,\ b=2,\ c=5\).
2. \(D=2^2-4(1)(5)\).
3. \(D=4-20=-16\).
4. Since \(D<0\), there are no real roots.
5. Answer: \(\boxed{\text{No real roots}}\).

6. Relationship Between Roots and Coefficients

Important Formula

If \(\alpha,\beta\) are the roots of \(ax^2+bx+c=0\), then

\[ \boxed{\alpha+\beta=-\frac ba} \qquad \boxed{\alpha\beta=\frac ca} \]

Solved Example 1 — Find sum and product

Question: For \(3x^2-8x+2=0\), find the sum and product of roots.

1. Compare with \(ax^2+bx+c=0\).
2. So \(a=3,\ b=-8,\ c=2\).
3. Sum \(=-\frac ba=-\frac{-8}{3}=\frac83\).
4. Product \(=\frac ca=\frac23\).
5. Answer: Sum \(=\boxed{\frac83}\), product \(=\boxed{\frac23}\).

Solved Example 2 — Find an unknown value

Question: If the roots of \(x^2-kx+12=0\) have sum \(7\), find \(k\).

1. Here \(a=1,\ b=-k,\ c=12\).
2. Sum of roots \(=-b/a\).
3. \(7=-\frac{-k}{1}=k\).
4. Therefore \(k=7\).
5. Answer: \(\boxed{k=7}\).

Solved Example 3 — Find an expression

Question: If \(\alpha,\beta\) are roots of \(x^2-9x+14=0\), find \(\alpha^2+\beta^2\).

1. Sum of roots: \(\alpha+\beta=9\).
2. Product of roots: \(\alpha\beta=14\).
3. Use \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\).
4. \(=9^2-2(14)\).
5. \(=81-28=53\).
6. Answer: \(\boxed{53}\).

7. Word Problems Based on Quadratic Equations

Basic Approach

In word problems, first choose a variable, form the equation from the information given, solve it, and finally reject any value that is not meaningful for the situation.

Solved Example 1 — Consecutive integers

Question: The product of two consecutive positive integers is \(56\). Find the integers.

1. Let the first integer be \(x\).
2. The next consecutive integer is \(x+1\).
3. Product is \(56\), so \(x(x+1)=56\).
4. \(x^2+x-56=0\).
5. Factorise: \((x+8)(x-7)=0\).
6. So \(x=-8\) or \(x=7\).
7. We need positive integers, so \(x=7\).
8. The next integer is \(8\).
9. Answer: \(\boxed{7,8}\).

Solved Example 2 — Rectangle

Question: The area of a rectangle is \(48\text{ cm}^2\). Its length is \(2\) cm more than its breadth. Find its dimensions.

1. Let breadth \(=x\) cm.
2. Then length \(=x+2\) cm.
3. Area \(=\text{length}\times\text{breadth}\).
4. So \(x(x+2)=48\).
5. \(x^2+2x-48=0\).
6. Factorise: \((x+8)(x-6)=0\).
7. So \(x=-8\) or \(x=6\).
8. Length and breadth cannot be negative, so breadth \(=6\) cm.
9. Length \(=6+2=8\) cm.
10. Answer: \(\boxed{6\text{ cm},8\text{ cm}}\).

Solved Example 3 — Speed and time

Question: A car covers \(120\) km. If its speed is increased by \(10\) km/h, it takes \(1\) hour less. Find the original speed.

1. Let original speed be \(x\) km/h.
2. Original time \(=\frac{120}{x}\) hours.
3. New speed \(=x+10\) km/h.
4. New time \(=\frac{120}{x+10}\) hours.
5. According to the question, \(\frac{120}{x}-\frac{120}{x+10}=1\).
6. Multiply by \(x(x+10)\): \(120(x+10)-120x=x(x+10)\).
7. \(1200=x^2+10x\).
8. So \(x^2+10x-1200=0\).
9. Factorise: \((x+40)(x-30)=0\).
10. Thus \(x=-40\) or \(x=30\).
11. Speed cannot be negative, so \(x=30\) km/h.
12. Answer: Original speed \(=\boxed{30\text{ km/h}}\).

8. Fully Solved Board-Style Questions

Question 1. Solve \(2x^2-5x+2=0\).

Solution:

\[2x^2-5x+2=(2x-1)(x-2)\]

\[(2x-1)(x-2)=0\]

Therefore \(2x-1=0\) or \(x-2=0\).

\[x=\frac12\quad\text{or}\quad x=2\]

Answer: \(\boxed{x=\frac12,2}\).

Question 2. Solve \(3x^2+2x-1=0\).

Solution:

Here \(a=3,\ b=2,\ c=-1\).

\[ x=\frac{-2\pm\sqrt{2^2-4(3)(-1)}}{6} \]

\[ x=\frac{-2\pm\sqrt{16}}6 =\frac{-2\pm4}{6} \]

Thus \(x=\frac13\) or \(x=-1\).

Answer: \(\boxed{x=\frac13,-1}\).

Question 3. Find the nature of roots of \(5x^2-6x+2=0\).

Solution:

Here \(a=5,\ b=-6,\ c=2\).

\[ D=(-6)^2-4(5)(2)=36-40=-4 \]

Since \(D<0\), the equation has no real roots.

Answer: \(\boxed{\text{No real roots}}\).

Question 4. If one root of \(2x^2+kx+3=0\) is \(1\), find \(k\).

Solution:

Since \(1\) is a root, substitute \(x=1\).

\[ 2(1)^2+k(1)+3=0 \]

\[ 2+k+3=0 \]

\[ k=-5 \]

Answer: \(\boxed{k=-5}\).

Question 5. Find the value of \(k\) for which \(x^2-2(k+1)x+k^2=0\) has equal roots.

Solution:

For equal roots, \(D=0\).

\[ [-2(k+1)]^2-4(1)(k^2)=0 \]

\[ 4(k+1)^2-4k^2=0 \]

\[ (k+1)^2-k^2=0 \]

\[ 2k+1=0 \]

\[ \boxed{k=-\frac12} \]

9. How to Write Quadratic Equation Answers in the Board Exam

1. First write the given equation clearly.
2. Convert it into \(ax^2+bx+c=0\) if needed.
3. Write \(a,b,c\) before using the quadratic formula.
4. Write the formula before substitution.
5. Show the discriminant calculation clearly.
6. Write both roots separately when using \(\pm\).
7. In word problems, reject impossible values with a reason.
8. Put the final answer in a box.

Common Mistakes to Avoid

• Forgetting to make the equation equal to zero.
• Writing the wrong sign for \(b\).
• Forgetting the square on \(b\) in \(b^2-4ac\).
• Using \(2a\) incorrectly as \(2+a\).
• Forgetting the \(\pm\) sign in the quadratic formula.
• Accepting a negative value when the question asks for length, speed or a positive integer.

10. Final 96% Target Revision

Must-Remember Formulas

\[\boxed{ax^2+bx+c=0,\quad a\neq0}\]

\[\boxed{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\]

\[\boxed{D=b^2-4ac}\]

\[ \boxed{\alpha+\beta=-\frac ba} \qquad \boxed{\alpha\beta=\frac ca} \]

Nature of roots: \(D>0\) → two distinct real roots; \(D=0\) → two equal real roots; \(D<0\) → no real roots.

One-Minute Checklist

• ✓ Know the standard form \(ax^2+bx+c=0\).
• ✓ Identify \(a,b,c\) correctly.
• ✓ Know how to solve by factorisation.
• ✓ Know the quadratic formula.
• ✓ Calculate the discriminant carefully.
• ✓ Identify the nature of roots.
• ✓ Use root-coefficient relationships correctly.
• ✓ Solve word problems step by step.
• ✓ Show complete working in board answers.

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Class 10 Maths • Chapter 4: Quadratic Equations

Designed as easy-to-read handwritten-style board exam notes.