Comprehensive Study Guide: Quadratic Equations

Historical Background and Development of Quadratic Equations

• Babylonians: Believed to be the first to solve quadratic equations. They knew how to find two positive numbers with a given positive sum and a given positive product, a problem mathematically equivalent to solving a quadratic equation of the form $x^2 - px + q = 0$.

• Euclid (Greek Mathematician): Developed a geometrical approach for finding lengths which correspond to solutions of quadratic equations.

• Ancient Indian Mathematicians: Credited with solving quadratic equations in general form.

  • Brahmagupta (C.E. 598–665): Provided an explicit formula to solve a quadratic equation of the form $ax^2 + bx = c$.

  • Sridharacharya (C.E. 1025): Derived a formula, known as the quadratic formula (as quoted by Bhaskara II), for solving a quadratic equation using the method of completing the square.

• Al-Khwarizmi (Arab Mathematician, C.E. 800): Studied quadratic equations of various distinct types.

• Abraham bar Hiyya Ha-Nasi: Published the book Liber embadorum in Europe in C.E. 1145, providing complete solutions of different quadratic equations.

Definition and Standard Form of Quadratic Equations

• Definition: A quadratic equation in the variable $x$ is an equation of the form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are real numbers, and $a \neq 0$.

• Polynomial Connection: Any equation of the form $p(x) = 0$, where $p(x)$ is a polynomial of degree $2$, is a quadratic equation.

• Standard Form: When the terms of $p(x)$ are written in descending order of their degrees, the equation is in standard form:   $ax^2 + bx + c = 0, \quad a \neq 0$

• Examples of Quadratic Equations:

  • $2x^2 + x - 300 = 0$

  • $2x^2 - 3x + 1 = 0$

  • $4x - 3x^2 + 2 = 0$

  • $1 - x^2 + 300 = 0$

Mathematical Modeling with Quadratic Equations

• Carpet Area of a Prayer Hall:

  • A charity trust decides to build a prayer hall with a carpet area of $300\,\text{m}^2$.

  • The length of the hall is one metre more than twice its breadth.

  • Let the breadth of the hall be $x\,\text{m}$. Then length is $(2x + 1)\,\text{m}$.

  • $\text{Area} = (2x + 1) \cdot x\,\text{m}^2 = (2x^2 + x)\,\text{m}^2$

  • Equating to given area: $2x^2 + x = 300 \implies 2x^2 + x - 300 = 0$

  • The breadth $x$ satisfies the quadratic equation $2x^2 + x - 300 = 0$

• Marble Problem (John and Jivanti):

  • John and Jivanti together have $45$ marbles. Both lost $5$ marbles each, and the product of the number of marbles they now have is $124$.

  • Let the number of marbles John had initially be $x$. Then Jivanti had $45 - x$.

  • Marbles remaining with John after losing $5$: $x - 5$

  • Marbles remaining with Jivanti after losing $5$: $45 - x - 5 = 40 - x$

  • Product of remaining marbles: $(x - 5)(40 - x) = 40x - x^2 - 200 + 5x = -x^2 + 45x - 200$

  • Given product is $124$: $-x^2 + 45x - 200 = 124 \implies -x^2 + 45x - 324 = 0 \implies x^2 - 45x + 324 = 0$

  • The number of marbles John had satisfies $x^2 - 45x + 324 = 0$

• Toy Production Problem (Cottage Industry):

  • A cottage industry produces a certain number of toys in a day ($x$).

  • The cost of production of each toy (in rupees) is $55$ minus the number of toys produced in a day: $(55 - x)$.

  • On a particular day, total cost of production is 750 rupees.

  • $\text{Total Cost} = x(55 - x) = 55x - x^2$

  • Equating total cost: $55x - x^2 = 750 \implies -x^2 + 55x - 750 = 0 \implies x^2 - 55x + 750 = 0$

  • The number of toys produced that day satisfies $x^2 - 55x + 750 = 0$

Verification and Simplification of Quadratic Equations

• Important Caveat: An algebraic expression must be simplified into polynomial form before determining whether it is a quadratic equation.

• Verification Examples:

  • Example 2(i): Check $(x - 2)^2 + 1 = 2x - 3$

    • $\text{LHS} = (x - 2)^2 + 1 = x^2 - 4x + 4 + 1 = x^2 - 4x + 5$

    • Equation becomes: $x^2 - 4x + 5 = 2x - 3 \implies x^2 - 6x + 8 = 0$

    • This is of the form $ax^2 + bx + c = 0$. It is a quadratic equation.

  • Example 2(ii): Check $x(x + 1) + 8 = (x + 2)(x - 2)$

    • $\text{LHS} = x(x + 1) + 8 = x^2 + x + 8$

    • $\text{RHS} = (x + 2)(x - 2) = x^2 - 4$

    • Equation becomes: $x^2 + x + 8 = x^2 - 4 \implies x + 12 = 0$

    • This is not of the form $ax^2 + bx + c = 0$ (degree is $1$). It is not a quadratic equation.

  • Example 2(iii): Check $x(2x + 3) = x^2 + 1$

    • $\text{LHS} = 2x^2 + 3x$

    • Equation becomes: $2x^2 + 3x = x^2 + 1 \implies x^2 + 3x - 1 = 0$

    • This is of the form $ax^2 + bx + c = 0$. It is a quadratic equation.

  • Example 2(iv): Check $(x + 2)^3 = x^3 - 4$

    • $\text{LHS} = (x + 2)^3 = x^3 + 6x^2 + 12x + 8$

    • Equation becomes: $x^3 + 6x^2 + 12x + 8 = x^3 - 4 \implies 6x^2 + 12x + 12 = 0 \implies x^2 + 2x + 2 = 0$

    • The cubic terms cancel out. This is of the form $ax^2 + bx + c = 0$. It is a quadratic equation.

Solution of Quadratic Equations by Factorisation

• Roots and Zeroes Concept:

  • A real number $\alpha$ is called a root of the quadratic equation $ax^2 + bx + c = 0, a \neq 0$ if $a\alpha^2 + b\alpha + c = 0$.

  • $x = \alpha$ is a solution of the quadratic equation, or $\alpha$ satisfies the quadratic equation.

  • The zeroes of the quadratic polynomial $ax^2 + bx + c$ and the roots of the quadratic equation $ax^2 + bx + c = 0$ are identical.

  • A quadratic equation can have at most two real roots.

• Factorisation Technique:

  • Factorise $ax^2 + bx + c$ into a product of two linear factors by splitting the middle term.

  • Equate each linear factor to zero to obtain the roots.

• Worked Examples:

  • Example 3: Find roots of $2x^2 - 5x + 3 = 0$

    • Split middle term $-5x$ as $-2x - 3x$ (since $(-2x) \times (-3x) = 6x^2 = (2x^2) \times 3$).

    • $2x^2 - 2x - 3x + 3 = 2x(x - 1) - 3(x - 1) = (2x - 3)(x - 1) = 0$

    • $2x - 3 = 0 \implies x = \frac{3}{2}$, or $x - 1 = 0 \implies x = 1$

    • Roots are $1$ and $\frac{3}{2}$.

  • Example 4: Find roots of $6x^2 - x - 2 = 0$

    • Split middle term $-x$ as $+3x - 4x$

    • $6x^2 + 3x - 4x - 2 = 3x(2x + 1) - 2(2x + 1) = (3x - 2)(2x + 1) = 0$

    • $3x - 2 = 0 \implies x = \frac{2}{3}$, or $2x + 1 = 0 \implies x = -\frac{1}{2}$

    • Roots are $\frac{2}{3}$ and $-\frac{1}{2}$.

  • Example 5: Find roots of $3x^2 - 2\sqrt{6}x + 2 = 0$

    • $3x^2 - \sqrt{6}x - \sqrt{6}x + 2 = \sqrt{3}x(\sqrt{3}x - \sqrt{2}) - \sqrt{2}(\sqrt{3}x - \sqrt{2}) = (\sqrt{3}x - \sqrt{2})(\sqrt{3}x - \sqrt{2}) = 0$

    • $\sqrt{3}x - \sqrt{2} = 0 \implies x = \sqrt{\frac{2}{3}}$

    • The root is repeated twice: $\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}$.

  • Example 6: Find dimensions of Prayer Hall ($2x^2 + x - 300 = 0$)

    • $2x^2 - 24x + 25x - 300 = 0 \implies 2x(x - 12) + 25(x - 12) = 0 \implies (x - 12)(2x + 25) = 0$

    • $x = 12$ or $x = -12.5$

    • Since breadth cannot be negative, breadth $x = 12\,\text{m}$.

    • Length $= 2x + 1 = 2(12) + 1 = 25\,\text{m}$.

Nature of Roots and the Discriminant

• Quadratic Formula:

  • The roots of $ax^2 + bx + c = 0$ are given by:   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

• Discriminant ($D$):

  • The term $b^2 - 4ac$ determines whether the quadratic equation has real roots or not, and is called the discriminant.

• Three Cases for Nature of Roots:

  1. Two distinct real roots: If $b^2 - 4ac > 0$, roots are $\frac{-b + \sqrt{b^2 - 4ac}}{2a}$ and $\frac{-b - \sqrt{b^2 - 4ac}}{2a}$.

  2. Two equal (coincident) real roots: If $b^2 - 4ac = 0$, roots are $-\frac{b}{2a}$ and $-\frac{b}{2a}$.

  3. No real roots: If $b^2 - 4ac < 0$, there is no real number whose square is negative, so no real roots exist.

• Worked Examples:

  • Example 7: Find discriminant of $2x^2 - 4x + 3 = 0$ and nature of roots.

    • $a = 2, b = -4, c = 3$

    • Discriminant $b^2 - 4ac = (-4)^2 - 4(2)(3) = 16 - 24 = -8 < 0$

    • Since discriminant is negative, there are no real roots.

  • Example 8: Pole erection on circular park boundary.

    • Diameter of circular park $AB = 13\,\text{m}$. Pole $P$ erected on boundary such that difference of distances from opposite gates $A$ and $B$ is $7\,\text{m}$.

    • Let $BP = x\,\text{m}$. Then $AP = (x + 7)\,\text{m}$.

    • Since $AB$ is diameter, $\angle APB = 90^\circ$ (angle in a semicircle).

    • By Pythagoras theorem: $AP^2 + PB^2 = AB^2 \implies (x + 7)^2 + x^2 = 13^2$

    • $x^2 + 14x + 49 + x^2 = 169 \implies 2x^2 + 14x - 120 = 0 \implies x^2 + 7x - 60 = 0$

    • Discriminant $b^2 - 4ac = 7^2 - 4(1)(-60) = 49 + 240 = 289 > 0$. Real roots exist, so pole can be erected.

    • Solving via quadratic formula: $x = \frac{-7 \pm \sqrt{289}}{2} = \frac{-7 \pm 17}{2}$

    • $x = 5$ or $x = -12$. Reject $x = -12$ (distance must be positive).

    • Distance from gate $B = 5\,\text{m}$, distance from gate $A = 12\,\text{m}$.

  • Example 9: Discriminant and roots of $3x^2 - 2x + \frac{1}{3} = 0$

    • $a = 3, b = -2, c = \frac{1}{3}$

    • Discriminant $b^2 - 4ac = (-2)^2 - 4(3)\left(\frac{1}{3}\right) = 4 - 4 = 0$

    • Has two equal real roots: $x = -\frac{b}{2a} = -\frac{-2}{2(3)} = \frac{1}{3}$

    • Roots are $\frac{1}{3}$ and $\frac{1}{3}$.

Complete Exercise Problems Reference

• Exercise 4.1:

  • 1. Check whether the following are quadratic equations:

    • (i) $(x + 1)^2 = 2(x - 3)$

    • (ii) $x^2 - 2x = (-2)(3 - x)$

    • (iii) $(x - 2)(x + 1) = (x - 1)(x + 3)$

    • (iv) $(x - 3)(2x + 1) = x(x + 5)$

    • (v) $(2x - 1)(x - 3) = (x + 5)(x - 1)$

    • (vi) $x^2 + 3x + 1 = (x - 2)^2$

    • (vii) $(x + 2)^3 = 2x(x^2 - 1)$

    • (viii) $x^3 - 4x^2 - x + 1 = (x - 2)^3$

  • 2. Represent the following situations in the form of quadratic equations:

    • (i) Area of rectangular plot is $528\,\text{m}^2$. Length of plot is one more than twice its breadth. Find length and breadth.

    • (ii) Product of two consecutive positive integers is $306$. Find the integers.

    • (iii) Rohan's mother is $26\,\text{years}$ older than him. Product of their ages $3\,\text{years}$ from now will be $360$. Find Rohan's present age.

    • (iv) A train travels a distance of $480\,\text{km}$ at uniform speed. If speed had been $8\,\text{km/h}$ less, it takes $3\,\text{hours}$ more to cover the same distance. Find speed of train.

• Exercise 4.2:

  • 1. Find roots of following quadratic equations by factorisation:

    • (i) $x^2 - 3x - 10 = 0$

    • (ii) $2x^2 + x - 6 = 0$

    • (iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$

    • (iv) $2x^2 - x + \frac{1}{8} = 0$

    • (v) $100x^2 - 20x + 1 = 0$

  • 2. Solve the problems given in Example 1.

  • 3. Find two numbers whose sum is $27$ and product is $182$.

  • 4. Find two consecutive positive integers, sum of whose squares is $365$.

  • 5. The altitude of a right triangle is $7\,\text{cm}$ less than its base. If hypotenuse is $13\,\text{cm}$, find the other two sides.

  • 6. Cottage industry pottery articles: Cost of production per article (in rupees) was $3$ more than twice number of articles produced on that day. Total cost on that day was 90 rupees. Find number of articles produced and cost of each article.

• Exercise 4.3:

  • 1. Find nature of roots of following quadratic equations. If real roots exist, find them:

    • (i) $2x^2 - 3x + 5 = 0$

    • (ii) $3x^2 - 4\sqrt{3}x + 4 = 0$

    • (iii) $2x^2 - 6x + 3 = 0$

  • 2. Find values of $k$ for each equation so that they have two equal roots:

    • (i) $2x^2 + kx + 3 = 0$

    • (ii) $kx(x - 2) + 6 = 0$

  • 3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and area is $800\,\text{m}^2$? If so, find length and breadth.

  • 4. Is the following situation possible? If so, determine present ages: Sum of ages of two friends is $20\,\text{years}$. Four years ago, product of their ages in years was $48$.

  • 5. Is it possible to design a rectangular park of perimeter $80\,\text{m}$ and area $400\,\text{m}^2$? If so, find length and breadth.

Chapter Summary Points

• 1. A quadratic equation in variable $x$ is of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$.

• 2. A real number $\alpha$ is a root of $ax^2 + bx + c = 0$ if $a\alpha^2 + b\alpha + c = 0$. Zeroes of $ax^2 + bx + c$ and roots of $ax^2 + bx + c = 0$ are identical.

• 3. If $ax^2 + bx + c$ ($a \neq 0$) can be factorised into two linear factors, roots are found by equating each factor to zero.

• 4. Quadratic Formula: Roots of $ax^2 + bx + c = 0$ are given by $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, provided $b^2 - 4ac \ge 0$

• 5. Nature of Roots:

  • Two distinct real roots, if $b^2 - 4ac > 0$

  • Two equal roots (coincident roots), if $b^2 - 4ac = 0$

  • No real roots, if $b^2 - 4ac < 0$