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CBSE · Class 10 · 🧮 Maths · Chapter 4

QUADRATIC EQUATIONS

Standard form of a quadratic equationRoots of a quadratic equationFactorisation methodQuadratic formulaDiscriminant and nature of roots

Chapter 4, 'Quadratic Equations', introduces students to equations of the form ax² + bx + c = 0. It covers methods to solve these equations, such as factorisation and the quadratic formula, and explains the concept of the discriminant to determine the nature of the roots. Understanding quadratic equations is fundamental for higher mathematics and has applications in various real-world scenarios, from physics to engineering.

Quadratic Equation: Pehchaan aur Standard Form

Quadratic Equation ek polynomial equation hoti hai jismein variable ki highest power 2 hoti hai.

  • Standard Form: \(ax^2 + bx + c = 0\), jahaan:
  • \(a, b, c\) real numbers hain.
  • \(a \neq 0\) (agar \(a=0\) ho gaya toh linear equation ban jaayegi).
  • \(x\) variable hai.
  • Pehchaan: Koi bhi equation quadratic hai ya nahi, yeh check karne ke liye, sabhi terms ko ek side le aao aur simplify karo. Agar highest power 2 bachti hai, toh woh quadratic hai.
  • Example:
  • \(x^2 - 5x + 6 = 0\) (Quadratic hai, \(a=1, b=-5, c=6\))
  • \((x+1)^2 = 2(x-3)\) \(\Rightarrow x^2 + 2x + 1 = 2x - 6 \Rightarrow x^2 + 7 = 0\) (Quadratic hai, \(a=1, b=0, c=7\))
  • \((x-2)(x+1) = (x-1)(x+3)\) \(\Rightarrow x^2 - x - 2 = x^2 + 2x - 3 \Rightarrow -x - 2 = 2x - 3 \Rightarrow 3x - 1 = 0\) (Quadratic nahi hai, linear hai)
  • Roots/Solutions: Variable \(x\) ki woh values jo equation ko satisfy karti hain (LHS = RHS bana deti hain) unhe roots ya solutions kehte hain. Quadratic equation ke at most do real roots ho sakte hain.
❗Important

Zeroes of a quadratic polynomial \(ax^2 + bx + c\) aur roots of the quadratic equation \(ax^2 + bx + c = 0\) ek hi cheez hai.

🚧Misconception

Students aksar \(a=0\) ko bhool jaate hain check karna. Agar \(a=0\) hai, toh woh quadratic equation nahi hai.

Quadratic Equations ko Solve karne ke Methods

Quadratic equations ko solve karne ke teen main methods hain:

  1. Factorisation Method (Middle Term Splitting)
  2. Completing the Square Method
  3. Quadratic Formula Method

1. Factorisation Method (Middle Term Splitting)

  • Concept: Agar ek quadratic polynomial \(ax^2 + bx + c\) ko do linear factors \((px+q)(rx+s)\) mein factorise kiya jaa sakta hai, toh roots find karne ke liye har factor ko zero ke equal rakhte hain.
  • Steps:
  1. Equation ko standard form \(ax^2 + bx + c = 0\) mein likho.
  2. Middle term \(bx\) ko do terms mein split karo, say \(p_1x\) aur \(p_2x\), aise ki:
  • \(p_1 + p_2 = b\)
  • \(p_1 \times p_2 = a \times c\)
  1. Terms ko group karo aur common factors nikaalo.
  2. Linear factors ko zero ke equal rakho aur \(x\) ki values nikaalo.
  • Example: Solve \(x^2 - 5x + 6 = 0\)
  • \(a=1, b=-5, c=6\)
  • \(a \times c = 1 \times 6 = 6\)
  • \(b = -5\)
  • Do numbers jinka sum \(-5\) ho aur product \(6\) ho: \(-2\) aur \(-3\).
  • \(x^2 - 2x - 3x + 6 = 0\)
  • \(x(x-2) - 3(x-2) = 0\)
  • \((x-2)(x-3) = 0\)
  • \(x-2 = 0 \Rightarrow x = 2\)
  • \(x-3 = 0 \Rightarrow x = 3\)
  • Roots hain \(2\) aur \(3\).

2. Completing the Square Method

  • Concept: Is method mein, quadratic equation ko \((x+k)^2 = d\) ya \((x-k)^2 = d\) ki form mein transform karte hain, taaki square root lekar \(x\) ki value nikaal sakein.
  • Steps:
  1. Equation ko standard form \(ax^2 + bx + c = 0\) mein likho.
  2. Agar \(a \neq 1\) hai, toh poori equation ko \(a\) se divide karo taaki \(x^2\) ka coefficient \(1\) ho jaaye: \(x^2 + \frac{b}{a}x + \frac{c}{a} = 0\).
  3. Constant term \(\frac{c}{a}\) ko RHS mein shift karo: \(x^2 + \frac{b}{a}x = -\frac{c}{a}\).
  4. \(x\) ke coefficient \(\frac{b}{a}\) ke half \(\left(\frac{b}{2a}\right)\) ka square dono sides add karo: \(x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2\).
  5. LHS ko \(\left(x + \frac{b}{2a}\right)^2\) ki form mein likho: \(\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}\).
  6. Dono sides ka square root lo: \(x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}\).
  7. \(x\) ke liye solve karo: \(x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}\).
  • Example: Solve \(2x^2 - 7x + 3 = 0\) by completing the square.
  • Divide by 2: \(x^2 - \frac{7}{2}x + \frac{3}{2} = 0\)
  • Shift constant: \(x^2 - \frac{7}{2}x = -\frac{3}{2}\)
  • Add \(\left(\frac{-7/2}{2}\right)^2 = \left(\frac{-7}{4}\right)^2 = \frac{49}{16}\) to both sides:

\(x^2 - \frac{7}{2}x + \frac{49}{16} = -\frac{3}{2} + \frac{49}{16}\)

  • \(\left(x - \frac{7}{4}\right)^2 = \frac{-24 + 49}{16}\)
  • \(\left(x - \frac{7}{4}\right)^2 = \frac{25}{16}\)
  • \(x - \frac{7}{4} = \pm \sqrt{\frac{25}{16}}\)
  • \(x - \frac{7}{4} = \pm \frac{5}{4}\)
  • Case 1: \(x - \frac{7}{4} = \frac{5}{4} \Rightarrow x = \frac{7}{4} + \frac{5}{4} = \frac{12}{4} = 3\)
  • Case 2: \(x - \frac{7}{4} = -\frac{5}{4} \Rightarrow x = \frac{7}{4} - \frac{5}{4} = \frac{2}{4} = \frac{1}{2}\)
  • Roots hain \(3\) aur \(\frac{1}{2}\).

3. Quadratic Formula Method

  • Concept: Yeh method completing the square ka hi generalisation hai. Directly roots nikalne ka formula provide karta hai.
  • Formula: Quadratic equation \(ax^2 + bx + c = 0\) ke roots diye jaate hain:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

  • Steps:
  1. Equation ko standard form \(ax^2 + bx + c = 0\) mein likho.
  2. \(a, b, c\) ki values identify karo.
  3. Values ko quadratic formula mein substitute karo aur solve karo.
  • Example: Solve \(2x^2 - 7x + 3 = 0\) using quadratic formula.
  • \(a=2, b=-7, c=3\)
  • \(x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)}\)
  • \(x = \frac{7 \pm \sqrt{49 - 24}}{4}\)
  • \(x = \frac{7 \pm \sqrt{25}}{4}\)
  • \(x = \frac{7 \pm 5}{4}\)
  • Case 1: \(x = \frac{7 + 5}{4} = \frac{12}{4} = 3\)
  • Case 2: \(x = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}\)
  • Roots hain \(3\) aur \(\frac{1}{2}\).
🧮Formula

Quadratic Formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

💡Tip

Agar question mein specific method mention na ho, toh quadratic formula sabse easy aur reliable method hai, especially jab factorisation difficult ho.

Nature of Roots: Discriminant ka Kamaal

Quadratic formula mein \(\sqrt{b^2 - 4ac}\) term roots ke nature ko determine karta hai. Is term \(b^2 - 4ac\) ko Discriminant kehte hain, aur ise \(D\) ya \(\Delta\) se denote karte hain.

  • Discriminant: \(D = b^2 - 4ac\)
  • Roots ka Nature:
  1. Agar \(D > 0\) (Positive):
  • Equation ke do distinct real roots honge.
  • Roots honge: \(x_1 = \frac{-b + \sqrt{D}}{2a}\) aur \(x_2 = \frac{-b - \sqrt{D}}{2a}\).
  • Agar \(D\) ek perfect square hai, toh roots rational honge. Agar nahi, toh irrational.
  1. Agar \(D = 0\):
  • Equation ke do equal real roots honge (coincident roots).
  • Roots honge: \(x_1 = x_2 = \frac{-b}{2a}\).
  1. Agar \(D < 0\) (Negative):
  • Equation ke koi real roots nahi honge. Roots imaginary honge (jo Class 11 mein padhoge).
  • Summary Table:

| Value of Discriminant (D) | Nature of Roots | | :------------------------ | :-------------------------------------------- | | \(D > 0\) | Two distinct real roots | | \(D = 0\) | Two equal real roots (coincident roots) | | \(D < 0\) | No real roots (Imaginary roots) |

  • Example: Find the nature of roots for \(2x^2 - 4x + 3 = 0\).
  • \(a=2, b=-4, c=3\)
  • \(D = b^2 - 4ac = (-4)^2 - 4(2)(3)\)
  • \(D = 16 - 24\)
  • \(D = -8\)
  • Since \(D < 0\), equation ke koi real roots nahi hain.
📖Definition

Discriminant (D): Quadratic equation \(ax^2 + bx + c = 0\) mein, \(b^2 - 4ac\) ko discriminant kehte hain. Yeh roots ke nature ko decide karta hai.

⭐Remember

Agar question mein 'real roots exist' bola hai, toh iska matlab hai \(D \ge 0\) (yaani \(D > 0\) ya \(D = 0\)).

Real-Life Problems aur Quadratic Equations

Bahut saari real-life situations ko quadratic equations mein convert karke solve kiya jaa sakta hai. Ismein word problems ko mathematical form mein change karna hota hai.

  • Steps to solve word problems:
  1. Problem ko dhyan se padho aur identify karo ki kya find karna hai.
  2. Unknown quantity (jo find karna hai) ko ek variable (jaise \(x\)) assume karo.
  3. Problem mein di gayi conditions ko use karke ek quadratic equation banao.
  4. Quadratic equation ko kisi bhi suitable method (factorisation, completing the square, quadratic formula) se solve karo.
  5. Obtained roots ko original problem ke context mein check karo. Kabhi-kabhi negative roots ya fractional roots real-life situation mein valid nahi hote (jaise age, length, number of articles).
  • Common Scenarios:
  • Area aur perimeter se related problems (rectangular plots, parks).
  • Speed, distance, time problems.
  • Age related problems.
  • Number related problems (consecutive integers, product of numbers).
  • Work and time problems (rarely, but possible).
  • Example: Ek rectangular plot ka area \(528 \text{ m}^2\) hai. Plot ki length uski breadth ke do guna se ek zyada hai. Length aur breadth find karo.
  • Step 1 & 2: Assume breadth \(= x\) meters.

Toh, length \(= (2x+1)\) meters.

  • Step 3: Area of rectangle = length \(\times\) breadth

\(x(2x+1) = 528\) \(2x^2 + x = 528\) \(2x^2 + x - 528 = 0\)

  • Step 4: Solve using quadratic formula (ya factorisation).

\(a=2, b=1, c=-528\) \(D = b^2 - 4ac = (1)^2 - 4(2)(-528) = 1 + 4224 = 4225\) \(\sqrt{D} = \sqrt{4225} = 65\) \(x = \frac{-1 \pm 65}{2(2)} = \frac{-1 \pm 65}{4}\) \(x_1 = \frac{-1 + 65}{4} = \frac{64}{4} = 16\) \(x_2 = \frac{-1 - 65}{4} = \frac{-66}{4} = -16.5\)

  • Step 5: Breadth negative nahi ho sakti, so \(x = 16\) is the valid solution.

Breadth \(= 16\) meters. Length \(= 2(16) + 1 = 32 + 1 = 33\) meters.

  • Check: Area \(= 16 \times 33 = 528 \text{ m}^2\). Correct.
💡Tip

Word problems mein units ka dhyaan rakho aur final answer mein units mention karna mat bhoolo. Negative solutions ko reject karte waqt reason zaroor likho.

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