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Class 10 . Mathematics . Mathematics . Quadratic Equations

Chapter 4: Quadratic Equations

Ch 4

Mathematics

Class 10

Lecture 1

Introduction & Quadratic Equations

Reference video · Lecture 1Introduction & Quadratic Equations

Meaning and Standard Form

Standard form
$$ax^2+bx+c=0,\qquad a\ne 0$$

Use this form to identify whether an equation is quadratic.

Worked example

Write \(4x-3x^2+2=0\) in standard form and identify its coefficients.

  1. Rewrite the equation in descending powers of the variable.
    $$4x-3x^2+2=0\quad\Longrightarrow\quad -3x^2+4x+2=0$$
  2. Compare with the standard form and identify the coefficients.
    $$a=-3,\qquad b=4,\qquad c=2$$
Answer: The standard form is \(-3x^2+4x+2=0\), so \(a=-3\), \(b=4\), and \(c=2\). Since \(a\ne0\), it is quadratic.

Forming Quadratic Equations from Situations

Rectangular prayer hall: breadth x m, length 2x+1 m, and area 300 m².
Rectangular prayer hall: breadth \(x\) m, length \(2x+1\) m, and area \(300\text{ m}^2\).
Area-based equation
$$\text{Area}=\text{length}\times\text{breadth}$$

For a rectangle, area equals length multiplied by breadth; substitute the expressions given in the situation.

Worked example

A rectangular hall has area \(300\text{ m}^2\) and its length is one metre more than twice its breadth. Form a quadratic equation for its breadth.

  1. Let the breadth be \(x\) metres; express the length using the given condition.
    $$\text{breadth}=x,\qquad \text{length}=2x+1$$
  2. Use the area condition to form an equation.
    $$x(2x+1)=300$$
  3. Expand and rearrange into standard form.
    $$2x^2+x=300\quad\Longrightarrow\quad 2x^2+x-300=0$$
Answer: The required equation is \(2x^2+x-300=0\).
Practice

Together, John and Jivanti initially have \(45\) marbles. Each loses \(5\), and the product of their remaining marbles is \(124\). If John initially had \(x\), form the quadratic equation.

Hint: Jivanti initially has \(45-x\) marbles.
Key step: \((x-5)(40-x)=124\)
Solution: Expand and rearrange: \((x-5)(40-x)=124\Rightarrow -x^2+45x-324=0\), hence \(x^2-45x+324=0\).
Practice

A workshop makes \(x\) toys in a day. Each toy costs Rs. \(55-x\) to produce, and the total cost is Rs. \(750\). Form the quadratic equation.

Hint: Total cost is number of toys multiplied by cost per toy.
Key step: \(x(55-x)=750\)
Solution: \(x(55-x)=750\Rightarrow 55x-x^2=750\Rightarrow x^2-55x+750=0\).

Checking Whether an Equation Is Quadratic

Classify an equation

  1. Expand brackets and simplify each side.
  2. Bring all terms to one side and combine like terms.
  3. Check whether the resulting equation has degree \(2\) and a non-zero quadratic coefficient.
    $$ax^2+bx+c=0,\qquad a\ne0$$
Examples: simplify before classifying
Given equationSimplified formQuadratic?
\((x-2)^2+1=2x-3\)\(x^2-6x+8=0\)Yes
\(x(x+1)+8=(x+2)(x-2)\)\(x+12=0\)No; it is linear
\(x(2x+3)=x^2+1\)\(x^2+3x-1=0\)Yes
\((x+2)^3=x^3-4\)\(x^2+2x+2=0\)Yes
Worked example

Check whether \((x+2)^3=x^3-4\) is a quadratic equation.

  1. Expand the cube on the left.
    $$(x+2)^3=x^3+6x^2+12x+8$$
  2. Cancel matching cubic terms and move all terms to one side.
    $$x^3+6x^2+12x+8=x^3-4\quad\Longrightarrow\quad 6x^2+12x+12=0$$
  3. Divide by the common factor and identify the degree.
    $$x^2+2x+2=0$$
Answer: It reduces to \(x^2+2x+2=0\), so it is quadratic.
Practice

Is \(x(x+1)+8=(x+2)(x-2)\) a quadratic equation?

Hint: Expand both sides and cancel matching terms.
Key step: \(x^2+x+8=x^2-4\)
Solution: \(x+12=0\), which is linear; therefore, it is not quadratic.
⚠ Common mistake

Do not classify an equation by its original appearance alone. Simplify first: terms of degree \(2\) or higher may cancel, changing the equation's final degree.

Quick recap

  • A quadratic equation has standard form \(ax^2+bx+c=0\), with \(a\ne0\).
  • The equation's degree must be checked after simplification.
  • To model a situation, define the unknown, write the relationships, and rearrange into standard form.
  • For rectangle problems, use area \(=\) length \(\times\) breadth.
  • Quadratic terms can cancel, while higher-degree terms can also cancel and leave a quadratic equation.
Lecture 2

Solution Of A Quadratic Equation By Factorisation

Reference video · Lecture 2Solution Of A Quadratic Equation By Factorisation

Roots and Factorisation

Definition

Root of a quadratic equation — A real number \(\alpha\) is a root of \(ax^2+bx+c=0\) if \(a\alpha^2+b\alpha+c=0\), where \(a\ne 0\).

Roots and zeroes

The roots of \(ax^2+bx+c=0\) are exactly the zeroes of the quadratic polynomial \(ax^2+bx+c\). A quadratic equation has at most two roots.

Zero-product property

If a product is zero, at least one factor is zero: \(AB=0\) implies \(A=0\) or \(B=0\). Use this after factorising the quadratic.

Splitting the middle term
$$m+n=b,\qquad mn=ac$$

For \(ax^2+bx+c\), choose \(m,n\) such that \(m+n=b\) and \(mn=ac\), then group and factor.

Solve by factorisation

  1. Write the equation in the form \(ax^2+bx+c=0\).
    $$ax^2+bx+c=0$$
  2. Find two terms whose sum is the middle-term coefficient and whose product is \(ac\); split the middle term.
    $$m+n=b,\qquad mn=ac$$
  3. Group terms and factorise into two linear factors.
    $$ax^2+bx+c=(px+q)(rx+s)$$
  4. Set each factor equal to zero and solve the resulting linear equations.
    $$(px+q)(rx+s)=0\Rightarrow px+q=0\text{ or }rx+s=0$$
Worked example

Find the roots of \(6x^2-x-2=0\) by factorisation.

  1. Solve \(6x^2-x-2=0\) by splitting the middle term.
    $$6x^2-x-2=0$$
  2. Split \(-x\) as \(3x-4x\), since the terms add to \(-x\) and their product is \(-12x^2\).
    $$6x^2+3x-4x-2=0$$
  3. Factor by grouping.
    $$3x(2x+1)-2(2x+1)=0\Rightarrow(3x-2)(2x+1)=0$$
  4. Set each factor to zero and solve.
    $$3x-2=0\Rightarrow x=\frac{2}{3},\qquad 2x+1=0\Rightarrow x=-\frac{1}{2}$$
Answer: The roots are \(\frac{2}{3}\) and \(-\frac{1}{2}\).
⚠ Common mistake

Do not divide by a factor that contains the variable; set each factor equal to zero so no root is lost.

Nature of Roots

Definition

Discriminant — For \(ax^2+bx+c=0\), the discriminant \(D=b^2-4ac\) determines the number and type of real roots.

Discriminant and nature of roots
DiscriminantNature of roots
\(D>0\)Two distinct real roots
\(D=0\)Two equal (coincident) real roots
\(D<0\)No real roots
Quadratic formula
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Use to find the roots directly; for real roots, the condition is \(D\ge 0\).

Determine the nature of roots

  1. Identify \(a\), \(b\), and \(c\) in standard form.
    $$ax^2+bx+c=0$$
  2. Calculate the discriminant.
    $$D=b^2-4ac$$
  3. Use the sign of \(D\) to classify the roots.
    $$D>0:\text{ distinct real};\quad D=0:\text{ equal real};\quad D<0:\text{ no real roots}$$
Worked example

Find the discriminant and nature of the roots of \(2x^2-4x+3=0\).

  1. Find the discriminant and nature of roots of \(2x^2-4x+3=0\).
    $$a=2,\quad b=-4,\quad c=3$$
  2. Calculate the discriminant.
    $$D=(-4)^2-4(2)(3)=16-24=-8$$
  3. Since the discriminant is negative, there are no real roots.
    $$D=-8<0$$
Answer: The discriminant is \(-8\), so the equation has no real roots.

Applying Quadratic Equations

Choosing a meaningful root

  • Applications: Form an equation from the situation, solve it, then reject roots that are not meaningful (for example, a negative distance or breadth).
  • Geometry: If a triangle is inscribed in a circle with one side as the diameter, the angle opposite the diameter is a right angle; Pythagoras’ theorem can then form the quadratic equation.
A triangle inscribed in a circle, with a diameter and labelled points.
A triangle inscribed in a circle, with a diameter and labelled points.
Practice

A hall has breadth \(x\) m, where \(2x^2+x-300=0\). Find its breadth and length if the length is \(2x+1\) m.

Hint: Factorise the quadratic and keep the positive value of the breadth.
Key step: \((x-12)(2x+25)=0\).
Solution: \(x=12\) or \(x=-12.5\). Reject the negative breadth, so the breadth is \(12\) m and the length is \(2(12)+1=25\) m.
Practice

A circular park has diameter \(13\) m. A pole on its boundary is \(7\) m farther from one of two diametrically opposite gates than from the other. Is this possible, and what are its distances from the gates?

Hint: Let the shorter distance be \(x\); use the right triangle formed by the two gates and the pole.
Key step: \((x+7)^2+x^2=13^2\), so \(x^2+7x-60=0\).
Solution: The discriminant is \(289\), and \(x=\frac{-7\pm17}{2}\) gives \(x=5\) or \(-12\). Reject the negative distance: it is possible, with distances \(5\) m and \(12\) m from the gates.
Practice

Find the nature and roots of \(3x^2-2\sqrt{6}x+2=0\).

Hint: Check the discriminant, then solve the repeated linear factor.
Key step: \(D=(-2\sqrt{6})^2-4(3)(2)=0\).
Solution: The roots are equal. Since \(3x^2-2\sqrt{6}x+2=(\sqrt{3}x-\sqrt{2})^2\), the repeated root is \(x=\sqrt{\frac{2}{3}}\).

Quick recap

  • A root makes the quadratic polynomial equal to zero; a quadratic has at most two roots.
  • Factorise the quadratic and set each linear factor equal to zero.
  • For middle-term splitting, choose terms with sum \(b\) and product \(ac\).
  • The discriminant is \(D=b^2-4ac\).
  • \(D>0\) gives two distinct real roots, \(D=0\) gives equal real roots, and \(D<0\) gives no real roots.
  • In applications, reject roots that cannot represent the quantity, such as negative lengths or distances.

Formula sheet

Standard form
$$ax^2+bx+c=0,\qquad a\ne 0$$
Area-based equation
$$\text{Area}=\text{length}\times\text{breadth}$$
Splitting the middle term
$$m+n=b,\qquad mn=ac$$
Quadratic formula
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Mathematics — ALL CHAPTERS

1

Real Numbers

2

Polynomials

3

Pair Of Linear Equations In Two Variables

4

Quadratic Equations

5

Arithmetic Progressions

6

Triangles

7

Coordinate Geometry

8

Introduction To Trigonometry

9

Some Applications Of Trigonometry

10

Circles

11

Areas Related To Circles

12

Surface areas And Volumes

13

Statistics 

14

Probability