Chapter 4 — QUADRATIC EQUATIONS

Concept Used: Discriminant Method

For a quadratic equation \(ax^2 + bx + c = 0\), the discriminant is defined as: \[ D = b^2 - 4ac \]

• If \(D > 0\): Two distinct real roots
• If \(D = 0\): Equal real roots
• If \(D < 0\): No real roots (complex roots)

Solution Roadmap (How to write in exam):

• Step 1: Identify \(a, b, c\)
• Step 2: Compute discriminant carefully
• Step 3: Conclude nature of roots
• Step 4: If real roots exist, apply quadratic formula

1. Given: \(2x^2 - 3x + 5 = 0\)

   Step 1: Identify coefficients

   \(a = 2,\; b = -3,\; c = 5\)

   Step 2: Compute discriminant

   \[ \begin{aligned} D &= b^2 - 4ac \\ &= (-3)^2 - 4 \times 2 \times 5 \\ &= 9 - 40 \\ &= -31 \end{aligned} \]

   Step 3: Nature of roots

   Since \(D < 0\), roots are non-real (complex).

   Final Answer:

   No real roots exist.

   Exam Insight: Negative discriminant confirms no x-intercept. Frequently asked in CBSE 1-mark MCQs and NTSE.

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2. Given: \(3x^2 - 4\sqrt{3}x + 4 = 0\)

   Step 1: Identify coefficients

   \(a = 3,\; b = -4\sqrt{3},\; c = 4\)

   Step 2: Compute discriminant

   \[ \begin{aligned} D &= (-4\sqrt{3})^2 - 4 \times 3 \times 4 \\ &= (16 \times 3) - 48 \\ &= 48 - 48 \\ &= 0 \end{aligned} \]

   Step 3: Nature of roots

   Since \(D = 0\), roots are equal and real.

   Step 4: Find roots

   \[ \begin{aligned} x &= \frac{-b}{2a} \\ &= \frac{-(-4\sqrt{3})}{2 \times 3} \\ &= \frac{4\sqrt{3}}{6} \end{aligned} \]

   Step 5: Simplify

   \[ \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3} \]

   Final Answer:

   Equal roots: \(x = \frac{2\sqrt{3}}{3}\)

   Exam Insight: Perfect square discriminant case. Common in CBSE 2-mark and algebra simplification problems.

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3. Given: \(2x^2 - 6x + 3 = 0\)

   Step 1: Identify coefficients

   \(a = 2,\; b = -6,\; c = 3\)

   Step 2: Compute discriminant

   \[ \begin{aligned} D &= (-6)^2 - 4 \times 2 \times 3 \\ &= 36 - 24 \\ &= 12 \end{aligned} \]

   Step 3: Nature of roots

   Since \(D > 0\), roots are real and distinct.

   Step 4: Apply quadratic formula

   \[ \begin{aligned} x &= \frac{-b \pm \sqrt{D}}{2a} \\ &= \frac{-(-6) \pm \sqrt{12}}{2 \times 2} \\ &= \frac{6 \pm \sqrt{12}}{4} \end{aligned} \]

   Step 5: Simplify root

   \[ \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \]

   \[ \begin{aligned} x &= \frac{6 \pm 2\sqrt{3}}{4} \\ &= \frac{3 \pm \sqrt{3}}{2} \end{aligned} \]

   Final Answer:

   \(x = \frac{3 + \sqrt{3}}{2}, \quad x = \frac{3 - \sqrt{3}}{2}\)

   Exam Insight: Simplifying surds is critical. Frequently appears in CBSE 3-mark and foundation for JEE algebra.

Concept: Condition for Equal Roots (Repeated Roots)

For a quadratic equation \(ax^2 + bx + c = 0\), the roots are equal when: \[ D = b^2 - 4ac = 0 \]

Important condition: The equation must remain quadratic (i.e., \(a \neq 0\)).

Solution Roadmap:

• Step 1: Convert equation into standard form
• Step 2: Identify \(a, b, c\)
• Step 3: Apply \(D = 0\)
• Step 4: Solve for parameter \(k\)
• Step 5: Check validity (quadratic condition)

1. Given: \(2x^2 + kx + 3 = 0\)

   Step 1: Identify coefficients

   \(a = 2,\; b = k,\; c = 3\)

   Step 2: Apply condition for equal roots

   \[ D = b^2 - 4ac = 0 \]

   Step 3: Substitute values

   \[ \begin{aligned} k^2 - 4(2)(3) &= 0 \\ k^2 - 24 &= 0 \end{aligned} \]

   Step 4: Solve equation

   \[ \begin{aligned} k^2 &= 24 \\ k &= \pm \sqrt{24} \end{aligned} \]

   Step 5: Simplify radical

   \[ \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6} \]

   \[ k = \pm 2\sqrt{6} \]

   Final Answer:

   \(k = 2\sqrt{6}\) or \(k = -2\sqrt{6}\)

   Exam Insight: Parameter-based discriminant problems are very common in CBSE 3-mark questions and form the base of higher algebra in JEE.

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2. Given: \(kx(x - 2) + 6 = 0\)

   Step 1: Convert into standard form

   \[ \begin{aligned} kx(x - 2) + 6 &= 0 \\ kx^2 - 2kx + 6 &= 0 \end{aligned} \]

   Step 2: Identify coefficients

   \(a = k,\; b = -2k,\; c = 6\)

   Step 3: Apply discriminant condition

   \[ D = b^2 - 4ac = 0 \]

   Step 4: Substitute values

   \[ \begin{aligned} (-2k)^2 - 4(k)(6) &= 0 \\ 4k^2 - 24k &= 0 \end{aligned} \]

   Step 5: Factorize

   \[ \begin{aligned} 4k^2 - 24k &= 0 \\ 4k(k - 6) &= 0 \end{aligned} \]

   Step 6: Solve

   \[ k = 0 \quad \text{or} \quad k = 6 \]

   Step 7: Check validity

   If \(k = 0\), then equation becomes: \[ 6 = 0 \] which is not a quadratic equation and has no solution.

   Therefore, \(k = 0\) is invalid.

   Final Answer:

   \(k = 6\)

   Exam Insight: Always check whether the equation remains quadratic. This is a common trap in CBSE and Olympiad questions.

Concept: Converting Word Problem into Quadratic Equation

Many real-life problems involving area, geometry, or ratios lead to quadratic equations. Here, we use: \[ \text{Area of rectangle} = \text{Length} \times \text{Breadth} \]

Solution Roadmap:

• Step 1: Assume one variable
• Step 2: Express other quantity using given condition
• Step 3: Form equation using area relation
• Step 4: Solve quadratic equation
• Step 5: Interpret solution (reject non-physical values)

Given:

• Length = twice breadth
• Area = \(800 \, m^2\)

Step 1: Assume variable

Let breadth = \(x\) m

Step 2: Express length

Length = \(2x\) m

Step 3: Form equation using area

\[ \text{Area} = \text{Length} \times \text{Breadth} \]

\[ x \times 2x = 800 \]

Step 4: Simplify equation

\[ \begin{aligned} 2x^2 &= 800 \\ x^2 &= 400 \end{aligned} \]

Step 5: Solve

\[ \begin{aligned} x &= \pm \sqrt{400} \\ &= \pm 20 \end{aligned} \]

Step 6: Interpret solution

Since dimensions represent physical lengths, they must be positive.

Therefore, reject \(x = -20\) and take: \[ x = 20 \]

Step 7: Find dimensions

Breadth = \(x = 20\) m
Length = \(2x = 2 \times 20 = 40\) m

Step 8: Verification

\[ \text{Area} = 40 \times 20 = 800 \, m^2 \]

Final Answer:

Yes, it is possible.
Length = \(40\) m
Breadth = \(20\) m

Concept: Age Problems → Formation of Quadratic Equation

In age-based problems:

• Present age is taken as a variable
• Past age = Present age − years
• Relationships (sum/product) are used to form equations

If the resulting quadratic equation has:

• Real solutions → situation is possible
• No real solution → situation is not possible

Solution Roadmap:

• Step 1: Assume present ages
• Step 2: Express past ages
• Step 3: Form equation using product condition
• Step 4: Simplify into quadratic form
• Step 5: Check discriminant
• Step 6: Interpret result

Given:

• Sum of present ages = 20
• Product of ages 4 years ago = 48

Step 1: Assume present ages

Let present age of first friend = \(x\) years

Then second friend's age = \(20 - x\) years

Step 2: Express past ages

Four years ago:
First friend = \(x - 4\)
Second friend = \((20 - x) - 4 = 16 - x\)

Step 3: Form equation

\[ (x - 4)(16 - x) = 48 \]

Step 4: Expand completely

\[ \begin{aligned} (x - 4)(16 - x) &= x(16 - x) - 4(16 - x) \\ &= 16x - x^2 - 64 + 4x \\ &= -x^2 + 20x - 64 \end{aligned} \]

So, \[ -x^2 + 20x - 64 = 48 \]

Step 5: Bring to standard form

\[ \begin{aligned} -x^2 + 20x - 64 - 48 &= 0 \\ -x^2 + 20x - 112 &= 0 \end{aligned} \]

Multiply entire equation by \(-1\):

\[ x^2 - 20x + 112 = 0 \]

Step 6: Compute discriminant

\[ \begin{aligned} D &= b^2 - 4ac \\ &= (-20)^2 - 4(1)(112) \\ &= 400 - 448 \\ &= -48 \end{aligned} \]

Step 7: Interpret result

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

Therefore, no real values of \(x\) satisfy the given condition.

Final Conclusion:

The given situation is not possible.

Concept: Rectangle with Fixed Perimeter → Quadratic Formation

For a rectangle:

• Perimeter \(= 2(l + b)\)
• Area \(= l \times b\)

When perimeter is fixed, one dimension can be expressed in terms of the other, leading to a quadratic equation.

Important insight: When discriminant \(D = 0\), length = breadth → rectangle becomes a square.

Solution Roadmap:

• Step 1: Use perimeter to relate \(l\) and \(b\)
• Step 2: Express one variable in terms of the other
• Step 3: Form area equation
• Step 4: Convert into quadratic form
• Step 5: Check discriminant
• Step 6: Interpret dimensions

Given:

• Perimeter = \(80\) m
• Area = \(400 \, m^2\)

Step 1: Use perimeter formula

\[ \begin{aligned} 2(l + b) &= 80 \\ l + b &= 40 \end{aligned} \]

Step 2: Assume variable

Let breadth = \(x\) m

Step 3: Express length

\[ l = 40 - x \]

Step 4: Form area equation

\[ \text{Area} = l \times b \]

\[ x(40 - x) = 400 \]

Step 5: Expand completely

\[ \begin{aligned} x(40 - x) &= 400 \\ 40x - x^2 &= 400 \end{aligned} \]

Step 6: Convert to standard quadratic form

\[ \begin{aligned} 40x - x^2 - 400 &= 0 \\ -x^2 + 40x - 400 &= 0 \end{aligned} \]

Multiply entire equation by \(-1\):

\[ x^2 - 40x + 400 = 0 \]

Step 7: Compute discriminant

\[ \begin{aligned} D &= b^2 - 4ac \\ &= (-40)^2 - 4(1)(400) \\ &= 1600 - 1600 \\ &= 0 \end{aligned} \]

Step 8: Nature of roots

Since \(D = 0\), roots are equal.

Step 9: Find value of \(x\)

\[ \begin{aligned} x &= \frac{-b}{2a} \\ &= \frac{-(-40)}{2 \times 1} \\ &= \frac{40}{2} \\ &= 20 \end{aligned} \]

Step 10: Find dimensions

Breadth = \(x = 20\) m
Length = \(40 - x = 40 - 20 = 20\) m

Step 11: Interpretation

Since length = breadth, the rectangle becomes a square.

Step 12: Verification

\[ \text{Perimeter} = 2(20 + 20) = 80 \]

\[ \text{Area} = 20 \times 20 = 400 \, m^2 \]

Final Answer:

Yes, it is possible.

Length = \(20\) m
Breadth = \(20\) m

The park is a square.