Quadratic Equations

• Linear equation in one variable & its solution:

Mathematical expression of equality in one variable with highest power 1, is known as linear equation in one variable.

Value of variable which satisfying given linear equations (i.e. LHS $=$ RHS) is known as solution of equation.

Eg. $7x-2=1+4x$ Put $\mathrm{x}=1$, LHS $=7(1)-2=5$ RHS $=1+4(1)=5$ so, $x=1$ is solution.

1.0Quadratic Equations

A quadratic equation in the variable $x$ is an equation of the form ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$, where $\mathrm{a},\mathrm{b},\mathrm{c}$ are real numbers, $\mathrm{a}\ne 0$. e.g., $2x^2-3x+1=0,4x-3x^2=0,1-x^2=0$ etc.

When we write the terms of a quadratic equation in descending order of their degrees, then we get the standard form of the equation i.e. $a{x}^2+bx+c=0$, where $a,b,c$ are real numbers and $a\ne 0$ is called the standard form of a quadratic equation.

2.0Types of quadratic equations

A quadratic equation can be of the following types : (i) $\mathrm{b}=0,\mathrm{c}\ne 0$ i.e. of the type ${\mathrm{ax}}^2+\mathrm{c}=0$ (Pure quadratic equation) (ii) $\mathrm{b}\ne 0,\mathrm{c}=0$ i.e. of the type ${\mathrm{ax}}^2+\mathrm{bx}=0$ (iii) $\mathrm{b}=0,\mathrm{c}=0$ i.e. of the type ${\mathrm{ax}}^2=0$ (iv) $\mathrm{b}\ne 0,\mathrm{c}\ne 0$ i.e. of the type ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$ (Mixed or complete quadratic equation)

Quadratic polynomial: A polynomial of the form ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}$, where $\mathrm{a},\mathrm{b}$ and c are real numbers and $\mathrm{a}\ne 0$ is called a quadratic polynomial. When we equate a quadratic polynomial to a constant, we get a quadratic equation.

Roots of quadratic equation: $x=\alpha$ is said to be root of the quadratic equation $a{x}^2+bx+c=0$, $a\ne 0$, if $x=\alpha$ satisfies the quadratic equation, i.e. in other words, the value of $a{\alpha }^2+b\alpha +c$ is zero.

• The zeroes of the quadratic polynomials $a{x}^2+bx+c$ and the roots of the quadratic equation are the same.

Bullding Concepts 1 Check whether the following are quadratic equations or not. (i) $(x+1{)}^2=2(x-3)$ (ii) $(x-2)(x+1)=(x-1)(x+3)$ Explanation: (i) Here, the given equation is $(x+1{)}^2=2(x-3)$ $\Rightarrow {\mathrm{x}}^2+2\mathrm{x}+1=2\mathrm{x}-6$ $\Rightarrow x^2+2\mathrm{x}-2\mathrm{x}+1+6=0$ $\Rightarrow x^2+7=0$ $\Rightarrow {\mathrm{x}}^2+0.\mathrm{x}+7=0$, which is of the form ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$ Hence, $(x+1{)}^2=2(x-3)$ is a quadratic equation. (ii) Here, the given equation is $(x-2)(x+1)=(x-1)(x+3)$ $\Rightarrow {\mathrm{x}}^2+\mathrm{x}-2\mathrm{x}-2={\mathrm{x}}^2+3\mathrm{x}-\mathrm{x}-3$ $\Rightarrow x^2-x^2-x-2x-2+3=0$ $\Rightarrow -3\mathrm{x}+1=0$, which is not of the form ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$ Hence, $(x-2)(x+1)=(x-1)(x+3)$ is not a quadratic equation

Numerical Abiity 1 Determine whether the given values are the solutions of the given equation or not. $\frac{2}{x^2}-\frac{5}{x}+2=0;x=5,x=\frac{1}{2}$ Solution: Putting $x=5$ and $x=\frac{1}{2}$ in the given equation $\frac{2}{(5{)}^2}-\frac{5}{5}+2$ and $\frac{2}{{\left(\frac{1}{2}\right)}^2}-\frac{5}{{\left(\frac{1}{2}\right)}^2}+2$ $\Rightarrow \frac{2}{25}-1+2$ and $\frac{2}{\frac{1}{4}}-\frac{5}{\frac{1}{2}}+2$ $\Rightarrow \frac{2}{25}+1$ and $8-10+2$ $\Rightarrow \frac{27}{25}$ and 0 i.e. $x=5$ does not satisfy but $x=\frac{1}{2}$ satisfies the given equation. Hence, $x=5$ is not a solution but $x=\frac{1}{2}$ is a solution of $\frac{2}{x^2}-\frac{5}{x}+2=0$.

3.0Solving Quadratic Equations through Factorisation

Given a quadratic equation ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0,\mathrm{a}\ne 0$.

Algorithm

Step-I: Factorise the product of constant term and coefficient of $x^2$ of the given quadratic equation.

Step-II : Express the coefficient of middle term as the sum or difference of the factors obtained in step - I. Clearly, the product of these two factors will be equal to the product of the coefficient of $x^2$ and constant term.

Step-III: Split the middle term in two parts obtained in step - II. Step-IV : Factorise the quadratic equation obtained in step - III by grouping method.

• Arrange the quadratic equation in standard form before applying factorisation of quadratic equation using middle term splitting.

Building Concepts 2 Solve the quadratic equation $6x^2-x-2=0$ by factorisation method. Explanation: $6x^2-x-2=0$ $\Rightarrow 6{\mathrm{x}}^2-4\mathrm{x}+3\mathrm{x}-2=0$ [We split middle term -x into $-4\mathrm{x}+3\mathrm{x}$ ] $\Rightarrow 2\mathrm{x}(3\mathrm{x}-2)+1(3\mathrm{x}-2)=0$ $\Rightarrow (2\mathrm{x}+1)(3\mathrm{x}-2)=0$ $\Rightarrow 2\mathrm{x}+1=0$ or $3\mathrm{x}-2=0$ $\Rightarrow \mathrm{x}=-\frac{1}{2}$ or $\mathrm{x}=\frac{2}{3}$

Numerical Ability 2 Solve the following quadratic equation by factorization method : $x^2-2ax+a^2-b^2=0$ Solution: Factors of the constant term ${\mathrm{a}}^2-{\mathrm{b}}^2$ are $(\mathrm{a}-\mathrm{b})\&(\mathrm{a}+\mathrm{b})$. Also, coefficient of the middle term $=-2\mathrm{a}$ $=-\{(\mathrm{a}-\mathrm{b})+(\mathrm{a}+\mathrm{b})\}$. $\Rightarrow {\mathrm{x}}^2-2\mathrm{ax}+{\mathrm{a}}^2-{\mathrm{b}}^2=0$ $\Rightarrow {\mathrm{x}}^2-\{(\mathrm{a}-\mathrm{b})+(\mathrm{a}+\mathrm{b})\}\mathrm{x}+(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})=0$ $\Rightarrow {\mathrm{x}}^2-(\mathrm{a}-\mathrm{b})\mathrm{x}-(\mathrm{a}+\mathrm{b})\mathrm{x}+(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})=0$ $\Rightarrow \mathrm{x}\{\mathrm{x}-(\mathrm{a}-\mathrm{b})\}-(\mathrm{a}+\mathrm{b})\{\mathrm{x}-(\mathrm{a}-\mathrm{b})\}=0$ $\Rightarrow \{\mathrm{x}-(\mathrm{a}-\mathrm{b})\}\{\mathrm{x}-(\mathrm{a}+\mathrm{b})\}=0$ $x-(a-b)=0$ or $x-(a+b)=0$ $\mathrm{x}=\mathrm{a}-\mathrm{b}$ or $\mathrm{x}=\mathrm{a}+\mathrm{b}$ OR $$\begin{gathered} \left(x^2-2ax+a^2\right)-b^2=0 \\ \Rightarrow (x-a{)}^2-(b{)}^2=0\left[\because x^2-2ax+a^2=(x-a{)}^2\right] \\ \Rightarrow (x-a+b)(x-a-b)=0\left[\because a^2-b^2=(a+b)(a-b)\right] \\ \Rightarrow x-(a-b)=0\text{ or }x-(a+b)=0 \\ \therefore x=a-b\text{ or }a+b \end{gathered}$$

• Two quadratic equations ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$ and $a^{\prime }{x}^2+b^{\prime }x+c^{\prime }=0$ have the same roots if and only if $\frac{\mathrm{a}}{{\mathrm{a}}^{\prime }}=\frac{\mathrm{b}}{{\mathrm{b}}^{\prime }}=\frac{\mathrm{c}}{{\mathrm{c}}^{\prime }}$
• When quadratic equation is in form of ${\mathrm{ax}}^2+\mathrm{bx}=0$ [where $\mathrm{c}=0$ ], then factorise directly by taken common from two terms.
• When quadratic equation is in form of $a{x}^2+c=0[$ where $b=0]$, then factorise by using algebraic identity ${\mathrm{a}}^2-{\mathrm{b}}^2=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})$.
• All the quadratic equation cannot be solved by factorisation. If factorisation is not applicable, then solve by other methods such as quadratic formula.

4.0Solving Quadratic Equations through Quadratic Formula (Sridharacharya formula)

Steps for derivation of quadratic formula:

Consider quadratic equation ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0,\mathrm{a}\ne 0$, then Step-1: Divide the equation by 'a' both sides $x^2+\frac{bx}{a}+\frac{c}{a}=\frac{0}{a}$

Step-2: Shift constant term to RHS. $x^2+\frac{bx}{a}=-\frac{c}{a}$

Step-3: Multiply the coefficient of $x$ by $\frac{1}{2}$ and add the square of term to both sides $x^2+\frac{b}{a}x+{\left(\frac{b}{2a}\right)}^2={\left(\frac{b}{2a}\right)}^2-\frac{c}{a}$

Step-4: Solve for x : $$\begin{gathered} {\left(x^2+\frac{b}{2a}\right)}^2=\frac{b^2}{4a^2}-\frac{c}{a} \\ {\left(x^2+\frac{b}{2a}\right)}^2=\frac{b^2-4ac}{4a^2} \\ {x}^2+\frac{b}{2a}=\sqrt{\frac{b^2-4ac}{4a^2}} \\ {x}^2=\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a} \\ {x}^2=\frac{\sqrt{b^2-4ac}}{2a}-\frac{b}{2a} \\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \end{gathered}$$ $\therefore$ The roots of the equation are $\mathrm{x}=\frac{-\mathrm{b}+\sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}$ or $\frac{-\mathrm{b}-\sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}$ Discriminant: If $a{x}^2+bx+c=0,a\ne 0(a,b,c\in R)$ is a quadratic equation, then the expression $b^2-4ac$ is known as its discriminant and is generally denoted by $D$ or $\Delta$. $\Rightarrow \mathrm{x}=\frac{-\mathrm{b}+\sqrt{\mathrm{D}}}{2\mathrm{a}}$ or $\mathrm{x}=\frac{-\mathrm{b}-\sqrt{\mathrm{D}}}{2\mathrm{a}}$, where $\mathrm{D}={\mathrm{b}}^2-4\mathrm{ac}$. Thus, if $D=b^2-4ac\ge 0$, then the quadratic equation $a{x}^2+bx+c=0$ has real roots $\alpha$ and $\beta$ given by $\alpha =\frac{-b+\sqrt{D}}{2a}$ and $\beta =\frac{-b-\sqrt{D}}{2a}$

• If $\mathrm{D}<0$, the roots are of the form $\mathrm{a}\pm \mathrm{ib}$ $(a,b\in R\&i=\sqrt{-1})$. If one root is $a+ib$, then the other root will be $\mathrm{a}-\mathrm{ib}$.

Numerical Ability 3 Solve $x^2-8x-20=0$ by using quadratic formula. Solution: On comparing the given equation ${\mathrm{x}}^2-8\mathrm{x}-20=0$ with ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$, we get $\mathrm{a}=1,\mathrm{b}=-8,\mathrm{c}=-20$ Hence, the required roots are $$\begin{gathered} x=\frac{-(-8)\pm \sqrt{(-8{)}^2-4\times 1\times (-20)}}{2\times 1}=\frac{8\pm \sqrt{64+80}}{2} \\ =\frac{8\pm \sqrt{144}}{2}=\frac{8\pm 12}{2}=\frac{8+12}{2}\text{ or }x=\frac{8-12}{2} \\ \Rightarrow x=10\text{ or }x=-2 \end{gathered}$$

Numerical Ability 4 Solve the quadratic equation $x^2-6x+4=0$ by using quadratic formula. (Sri Dharacharya's Rule). Solution: On comparing the given equation $x^2-6x+4=0$ with the standard form of quadratic equation $a{x}^2+bx+c=0$, we get $a=1,b=-6,c=4$. Hence the required roots are $x=\frac{-(-6)\pm \sqrt{(-6{)}^2-4(1)(4)}}{2(1)}$ $=\frac{6\pm \sqrt{36-16}}{2}$ $=\frac{6\pm \sqrt{20}}{2}$ $=\frac{6\pm \sqrt{4\times 5}}{2}$ $=\frac{2(3\pm \sqrt{5})}{2}$ $=3\pm \sqrt{5}$ Thus, the roots of the equation are $3+\sqrt{5}$ and $3-\sqrt{5}$.

5.0Nature of roots

Let the quadratic equation be $a{x}^2+bx+c=0$, where $a\ne 0$ and $a,b,c\in R$. The roots of the given equation are given by $x=\frac{-b\pm \sqrt{D}}{2a}$, where $D=b^2-4ac$ is the discriminant i.e. if $\alpha$ and $\beta$ are two roots of the quadratic equation. Then, $\alpha =\frac{-b+\sqrt{D}}{2a}$ and $\beta =\frac{-b-\sqrt{D}}{2a}$ Now, the following cases are possible.

Case-I: When D $>0$, Roots are real and unequal (distinct). The roots are given by $\alpha =\frac{-b+\sqrt{D}}{2a}$ and $\beta =\frac{-b-\sqrt{D}}{2a}$ Case-II: When D $=0$, Roots are real and equal and each root is given by $\alpha =\frac{-b}{2a}=\beta$ Case-III : When D $<0$, No real roots exist. Both the roots are imaginary.

• $D=b^2-4ac$ is discriminant not $\sqrt{D}$

• While solving quadratic equation by using formula, first find the value of $D$. If $D\ge 0$ then apply quadratic formula else no real roots.

Numerical Ability 5 Find the nature of the roots of the following equation. If the real roots exist, find them: $3x^2-7x+4=0$ Solution: The given equation $3{\mathrm{x}}^2-7\mathrm{x}+4=0$ comparing it with ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$, we get $a=3,b=-7$ and $c=4$. $\therefore$ Discriminant, $\mathrm{D}={\mathrm{b}}^2-4\mathrm{ac}$ $$\begin{gathered} =(-7{)}^2-4\times 3\times 4 \\ =49-48=1>0 \end{gathered}$$ $\because \mathrm{D}>0$, roots are real and unequal. Now, by quadratic formula, $x=\frac{-b\pm \sqrt{D}}{2a}=\frac{7\pm \sqrt{1}}{2\times 3}=\frac{7\pm 1}{6}$ Hence the roots are $\mathrm{x}=\frac{4}{3},1$

Numerical Ability 6 Find the value of $k$ for the quadratic equation $9x^2+8kx+16=0$, so that it has real and equal roots. Solution: The given equation is $9{\mathrm{x}}^2+8\mathrm{kx}+16=0$. Comparing it with ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$, we get $\mathrm{a}=9,\mathrm{b}=8\mathrm{k}$ and $\mathrm{c}=16$. $\therefore$ Discriminant, $\mathrm{D}={\mathrm{b}}^2-4\mathrm{ac}=(8\mathrm{k}{)}^2-4\times 9\times 16=64{\mathrm{k}}^2-576$ Since roots are real and equal, so $D=0$ $\Rightarrow 64{\mathrm{k}}^2-576=0$ $\Rightarrow 64{\mathrm{k}}^2=576$ $\Rightarrow {\mathrm{k}}^2=\frac{576}{64}=9$ $\Rightarrow \mathrm{k}=\pm 3$ Hence, $\mathrm{k}=3$ or -3

Numerical Ability 7 Find the set of values of $k$ for which the equation $k{x}^2+2x+1=0$ has distinct real roots. Solution: The given equation is ${\mathrm{kx}}^2+2\mathrm{x}+1=0$. $\therefore \mathrm{D}=(2{)}^2-4\times \mathrm{k}\times 1=4-4\mathrm{k}$. For distinct and real roots, we must have, $\mathrm{D}>0$. Now, D > 0 $\Rightarrow (4-4\mathrm{k})>0\Rightarrow 4>4\mathrm{k}\Rightarrow 4\mathrm{k}<4\Rightarrow \mathrm{k}<1$. But $\mathrm{k}\ne 0$ $\therefore$ All the values less than 1 except 0 is possible for k .

6.0Formation of quadratic equations using sum & product of roots

Let $\alpha$ and $\beta$ be the roots of the quadratic equation ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0,\mathrm{a}\ne 0$. Then $\alpha =\frac{-\mathrm{b}+\sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}$ and $\beta =\frac{-\mathrm{b}-\sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}$ $\therefore$ The sum of roots $\alpha +\beta =\frac{-b}{a}=-\frac{\text{ coefficient of }x}{\text{ coefficient of }{x}^2}$ and product of roots $=\alpha \beta =\frac{c}{a}=\frac{\text{ constant term }}{\text{ coefficient of }{x}^2}$ Hence the quadratic equation whose roots are $\alpha$ and $\beta$ is given by $x^2-(\alpha +\beta )x+\alpha \beta =0$ i.e. $x^2-($ sum of the roots $)x+$ product of the roots $=0$

Numerical Ability 8 Form the quadratic equation whose roots are $2+\sqrt{5}$ and $2-\sqrt{5}$. Solution: Here roots are $\alpha =2+\sqrt{5}$ and $\beta =2-\sqrt{5}$ $\therefore$ Sum of roots $=\alpha +\beta =(2+\sqrt{5})+(2-\sqrt{5})$ $\therefore \alpha +\beta =4$ and product of the roots $=\alpha \beta =(2+\sqrt{5})(2-\sqrt{5})=4-5=-1$ $\therefore \alpha \beta =-1$ $\therefore$ Required equation is ${\mathrm{x}}^2-$ (sum of roots) $\mathrm{x}+$ product of roots $=0$ or $x^2-(\alpha +\beta )x+\alpha \beta =0$ or $x^2-(4)x+(-1)=0$ $\therefore x^2-4\mathrm{x}-1=0$

7.0Problems on Numbers and Ages

Algorithm, the method of solving word problems consists of the following three steps Step-I: Translating the word problem into symbolic language (mathematical statement) which means identifying relationships existing in the problem and then forming the quadratic equation. Step-II: Solving the quadratic equation thus formed. Step-III: Interpreting the solution of the equation which means translating the result of mathematical statement into verbal language. Problems based on numbers

• If $a$ is unit digit of a number and $b$ is ten's digit of a number then the 2 digit number is $10b+a$.

Numerical Ability 9 The difference of two numbers is 3 and their product is 504 . Find the numbers. Solution Let the required numbers be $x$ and $(x-3)$. Then, $x(x-3)=504$ $\Rightarrow x^2-3x-504=0$ $\Rightarrow x^2-24x+21x-504=0$ $\Rightarrow x(x-24)+21(x-24)=0$ $\Rightarrow (x-24)(x+21)=0$ $\Rightarrow \mathrm{x}-24=0$ or $\mathrm{x}+21=0$ $\Rightarrow x=24$ or $x=-21$ If $x=-21$, then the numbers are -21 and -24 . Again, if $x=24$, then the numbers are 24 and 21. Hence, the numbers are $-21,-24$ or $24,21$.

8.0Problems based on ages

Numerical Ability 10 Seven years ago Varun's age was five times the square of Swati's age. Three years hence, Swati's age will be two-fifth of Varun's age. Find their present ages. Solution: Let the present age of Swati be x years. Then the age of Swati seven years ago was $\mathrm{x}-7$ years. According to question, seven years ago age of Varun was $5(x-7{)}^2$ Three years hence, Varun's age $=\left[5(x-7{)}^2+10\right]$ years and Swati's age $=(x+3)$ years. $\therefore (x+3)=\frac{2}{5}\left[5(x-7{)}^2+10\right]$ $\Rightarrow \mathrm{x}+3=\frac{2}{5}\times 5\left[(\mathrm{x}-7{)}^2+2\right]$ $\Rightarrow \mathrm{x}+3=2\left({\mathrm{x}}^2-14\mathrm{x}+49+2\right)$ $\Rightarrow \mathrm{x}+3=2{\mathrm{x}}^2-28\mathrm{x}+102$ $\Rightarrow 2x^2-29x+99=0$ $\Rightarrow 2{\mathrm{x}}^2-18\mathrm{x}-11\mathrm{x}+99=0$ $\Rightarrow 2\mathrm{x}(\mathrm{x}-9)-11(\mathrm{x}-9)=0$ $\Rightarrow (\mathrm{x}-9)(2\mathrm{x}-11)=0$ $\Rightarrow \mathrm{x}=9$ or $\mathrm{x}=\frac{11}{2}$ $\because \mathrm{x}=\frac{11}{2}$ is not possible $\left(\because \frac{11}{2}<7\right)$ So, $x=9$. Varun's ages 7 years ago were $5(x-7{)}^2=5(9-7{)}^2=5\times 4=20$ Therefore, Varun's present age $=20+7=27$ years Hence, Varun's present age is 27 years and Swati's present age is 9 years.

• While solving the problems based on ages, if age of any person is negative then ignore it because age cannot be negative.

9.0Problems on geometrical concepts

• Sum of angles of a triangle is $180^{\circ }$.
• Sum of angles of a quadrilateral is $360^{\circ }$.
• Area of triangle $=\frac{1}{2}\times$ base $\times$ height
• In $△\mathrm{ABC}$ right angled at $\mathrm{B},{\mathrm{AB}}^2+{\mathrm{BC}}^2={\mathrm{AC}}^2$
  

Numerical Ability 11 The length of the hypotenuse of a right triangle exceeds the length of the base by $2cm$ and exceeds twice the length of the altitude by 1 cm . Find the length of each side of the triangle. Solution:

• Sides and angles of triangle cannot be negative.

Additional Subtopics:

10.0Problems on speed, time and distance

• Distance $=$ speed $\times$ time

• 1 hour $=60$ minutes

• Let $\mathrm{x}\mathrm{km}/\mathrm{hr}$ be the speed of boat in still water and let $\mathrm{y}\mathrm{km}/\mathrm{hr}$ be speed of stream. Then speed upstream $=(x-y)\mathrm{km}/\mathrm{hr}$ speed downstream $=(x+y)km/hr$

Numerical Ability 12 A train travels 240 km at a uniform speed. If the speed had been $8\mathrm{km}/\mathrm{h}$ more, it would have taken 1 hour less for the same journey. Find the speed of the train. Solution: Let the speed of the train be $x\mathrm{km}/\mathrm{h}$. Then, Time taken to cover the distance of $240\mathrm{km}=\frac{240}{\mathrm{x}}$ hours. If the speed of the train is increased by $8\mathrm{km}/\mathrm{h}$. Then, Time taken to cover the same distance $=\left(\frac{240}{x+8}\right)\mathrm{h}$ According to the question, $$\begin{gathered} \frac{240}{x}-\frac{240}{x+8}=1 \\ \Rightarrow \frac{240(x+8)-240x}{x(x+8)}=1 \\ \Rightarrow 240x+1920-240x=x^2+8x \\ \Rightarrow x^2+8x-1920=0 \\ \Rightarrow x^2+48x-40x-1920=0 \\ \Rightarrow x(x+48)-40(x+48)=0 \\ \Rightarrow (x+48)(x-40)=0 \\ \Rightarrow x=-48 \\ \text{ or }x=40 \end{gathered}$$ But the speed cannot be negative. Hence, the speed of the train is $40\mathrm{km}/\mathrm{h}$.

• Speed can't be negative. So, negative speed is not considered as solution or final answer.

11.0Problems on time and work

• A alone takes $x$ days to complete some work.

Then, A's 1-day work $=\frac{1}{\mathrm{x}}$

Numerical Ability 13 A takes 9 days less than the time taken by $B$ to finish a piece of work. If both $A$ and $B$ together can finish it in $6$ days, find the time taken by $B$ to finish the work. Solution: Suppose B alone takes $x$ days to finish the work, then A alone can finish it in $(x-9)$ days. B's 1 day's work $=\frac{1}{x}$ A's 1 day's work $=\frac{1}{(x-9)}$ $(A+B{)}^{\prime }$ 's 1 day's work $=\frac{1}{6}$ $\therefore \frac{1}{x}+\frac{1}{(x-9)}=\frac{1}{6}$ $\Rightarrow \frac{(x-9)+x}{x(x-9)}=\frac{1}{6}$ $\Rightarrow \frac{(2x-9)}{\left(x^2-9x\right)}=\frac{1}{6}$ $\Rightarrow x^2-9x=12x-54$ $\Rightarrow x^2-21x+54=0$ $\Rightarrow x^2-18x-3x+54=0$ $\Rightarrow \mathrm{x}(\mathrm{x}-18)-3(\mathrm{x}-18)=0$ $\Rightarrow (\mathrm{x}-18)(\mathrm{x}-3)=0$ $\Rightarrow x-18=0$ or $x-3=0$ $\Rightarrow \mathrm{x}=18$ or $\mathrm{x}=3$ $\Rightarrow \mathrm{x}=18(\because \mathrm{x}=3$ not possible $)$ Hence, B alone can finish the work in 18 days.

• Number of days should be always positive. Ignore the negative value if coming while solving.

12.0Solution by completing the square

Algorithm

Step-I: Obtain the quadratic equation. Let the quadratic equation $bea{x}^2+bx+c=0,a\ne 0$.

Step-II: Make the coefficient of $x^2$ unity by dividing throughout by it, if it is not unity, i.e. obtain $x^2+\frac{b}{a}x+\frac{c}{a}=0$.

Step-III: Shift the constant term $\frac{c}{a}$ on RHS to get $x^2+\frac{b}{a}x=-\frac{c}{a}$

Step-IV : Add square of half of the coefficient of $x$, i.e. ${\left(\frac{b}{2a}\right)}^2$ on both sides to obtain $x^2+2\left(\frac{b}{2a}\right)x+{\left(\frac{b}{2a}\right)}^2={\left(\frac{b}{2a}\right)}^2-\frac{c}{a}$

Step-V : Write L.H.S. as the perfect square and simplify RHS to get ${\left(x+\frac{b}{2a}\right)}^2=\frac{b^2-4ac}{4a^2}$ Step-VI: Take square root of both sides to get $x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}$.

Step-VII : Obtain the values of $x$ by shifting the constant term $\frac{b}{2a}$ on RHS i.e. $x=-\frac{b}{2a}\pm \sqrt{\frac{b^2-4ac}{4a^2}}\Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Numerical Ability 14 Solve : $9{\mathrm{x}}^2-15\mathrm{x}+6=0$ Solution: Here, $9x^2-15x+6=0$ $$\begin{gathered} \Rightarrow x^2-\frac{15}{9}x+\frac{6}{9}=0 \\ \Rightarrow x^2-\frac{5}{3}x+\frac{2}{3}=0\text{ (Dividing throughout by }9\text{ ) } \end{gathered}$$ $\Rightarrow {\mathrm{x}}^2-\frac{5}{3}\mathrm{x}=-\frac{2}{3}$ (Shifting the constant term on RHS) $\Rightarrow x^2-2\left(\frac{5}{6}\right)x+{\left(\frac{-5}{6}\right)}^2$ $={\left(\frac{-5}{6}\right)}^2-\frac{2}{3}$ (Adding square of half of coefficient of $x$ on both sides) $\Rightarrow {\left(x-\frac{5}{6}\right)}^2=\frac{25}{36}-\frac{2}{3}$ $\Rightarrow {\left(x-\frac{5}{6}\right)}^2=\frac{25-24}{36}$ $\Rightarrow {\left(x-\frac{5}{6}\right)}^2=\frac{1}{36}$ $\Rightarrow \mathrm{x}-\frac{5}{6}=\pm \frac{1}{6}$ (Taking square root of both sides) $\Rightarrow \mathrm{x}=\frac{5}{6}\pm \frac{1}{6}$ $\Rightarrow x=\frac{5}{6}+\frac{1}{6}=1$ Or $\mathrm{x}=\frac{5}{6}-\frac{1}{6}=\frac{4}{6}=\frac{2}{3}$ $\Rightarrow \mathrm{x}=1$ or $\mathrm{x}=\frac{2}{3}$