Exercise Solutions
1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:(i) ${2x^2 – 7x + 3 = 0}$Solution Since the coefficient of $x^2$ is not a perfect square, multiply the equation throughout by its coefficient i.e 2 to make it a perfect square. i.e $4x^2-2×7x+6$ NOw consider $4x^2-2×7x$ part only, First make it a perfect square. Consider the $4x^2$ as a square with sides $2x$ and the $14x$ as a rectangle with sides $2x$ and $7$ (click the link for animation ) Now split the side of $7$ into 2 equal half.. i.e (7/2) $4x^2-2×7x$ => $(2x)^2 -(2x)(7/2) - (2x)(7/2) $ In order to make the square with sides $(2x-7/2)$ a perfect square, add a square of sides $7/2$ to this square and in order to balance the equation subract the square also . i.e Add and subract $(7/2)^2$ from above equation =$(2x)^2 -(2x)(7/2) - (2x)(7/2) +(7/2)^2 -(7/2)^2$ = $(2x)(2x-7/2) - (7/2){(2x-7/2)} - (7/2)^2 $ = $ (2x-7/2)(2x-7/2) - (7/2)^2 $ = $(2x-7/2)^2 - (7/2)^2 $ Therefore $4x^2-2×7x+6$ = ${[(2x-7/2)^2 - (7/2)^2 + 6] $ = ${[(2x-7/2)^2 - 49/4 + 6] $ = ${[(2x-7/2)^2 - (49-24)/4] $ = ${[(2x-7/2)^2 - 25/4] $ = ${[(2x-7/2)^2 - (5/2)^2] $ =$[2x-7/2+5/2][2x-7/2 - 5/2] $ =$(2x-1)(2x-6)$ Therefore the roots of the equations are => $(2x-1)=0$ and $(2x-6)=0$ => $x=1/2$ and $x=3$(ii) ${2x^2 + x – 4 = 0}$Solution Since the coefficient of $x^2$ is not a perfect square, multiply the equation throughout by its coefficient i.e 2 to make it a perfect square. Follow the procedure as described above. i.e ${4x^2 + 2x – 8 = 0}$ Now, $4x^2 + 2x$ = $4x^2 + 2x(1/2) + 2x(1/2)$ = $4x^2 + 2x(1/2) + 2x(1/2)+ (1/2)^2 - (1/2)^2 $ =$2x(2x+1/2) + 1/2(2x+1/2) - (1/2)^2 $ =$(2x+1/2)(2x+1/2) - (1/2)^2 $ =$(2x+1/2)^2 - (1/4) $ Therefore ${4x^2 + 2x – 8 }$ = $(2x+1/2)^2 - (1/4) - 8$ = $(2x+1/2)^2 -(1+32)/4$ = $(2x+1/2)^2 -(√33/2)^2$ = $[(2x+1/2) +(√33/2)][(2x+1/2) -(√33/2)]$ = $[(2x+ (1+√33)/2)][(2x+ (1-√33)/2 )]$ Therefore the roots of the equations are =>$(2x+ (1+√33)/2)=0$ and $(2x+ (1-√33)/2 )=0$ => $x=-(1+√33)/4 $ and $x=-(1-√33)/4$(iii) ${4x^2 +4√3x + 3 = 0}$Solution The coefficient of $x^2$ is a perfect square. DIvide the cofficent of $x$ into 2 equal halfs. Therefore $4x^2 +4√3x$ = $(2x)^2 +{4√3x}/2 +{4√3x}/2$ = $(2x)^2 +{(2x)√3} +{(2x)√3}$ Add and subract $(√3)^2$ from the equation. = $(2x)^2 +{(2x)√3} +{(2x)√3} + (√3)^2 -(√3)^2 $ =$2x(2x+ √3) +√3(2x+ √3) -(√3)^2 $ = $(2x+ √3)(2x+ √3)- (√3)^2$ = $(2x+ √3)^2- (√3)^2$ ${4x^2 +4√3x + 3 = 0}$ => $(2x+ √3)^2- (√3)^2 + 3$ = $(2x+ √3)^2 - 0^2$ Therefore the roots of the equations are => $2x+ √3 = 0$ and $2x+ √3=0$ => $x= -√3/2$ and $x= -√3/2$(iv) ${2x^2 + x + 4 = 0}$Solution Since the coefficient of $x^2$ is not a perfect square, multiply the equation throughout by its coefficient i.e 2 to make it a perfect square. Follow the procedure as described above. i.e ${4x^2 + 2x + 8 = 0}$ Now, $4x^2 + 2x$ = $4x^2 + 2x(1/2) + 2x(1/2)$ = $4x^2 + 2x(1/2) + 2x(1/2)+ (1/2)^2 - (1/2)^2 $ =$2x(2x+1/2) + 1/2(2x+1/2) - (1/2)^2 $ =$(2x+1/2)(2x+1/2) - (1/2)^2 $ =$(2x+1/2)^2 - (1/4) $ Therefore ${4x^2 + 2x + 8 }$ = $(2x+1/2)^2 - (1/4) + 8$ = $(2x+1/2)^2 -(1-32)/4$ = $(2x+1/2)^2 -(-31/4)$ = $(2x+1/2)^2 +(√31/2)^2$ Since, the equation cannot be solved, so the roots do not exist.2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.3. Find the roots of the following equations: (i) $x - 1/x=3 ,x ≠ 0$Solution $x - 1/x=3$ => $x^2-1=3x$ =>$x^2-3x-1$ Solving the eqn using quadratic formula, $x={-b±√{b^2-4ac}}/{2a}$ $x={-(-3)±√{9-4(1)(-1)}}/{2(1)}$ $x={3±√{13}}/{2}$ SO, the roots are ${3-√{13}}/{2}$ and ${3+√{13}}/{2}$(ii) ${1/{x+4} - 1/{x-7}=11/30},{, x ≠ – 4, 7}$Solution ${1/{x+4} - 1/{x-7}=11/30}$ => ${(x-7)-(x+4)}/{(x+4)(x-7)}$=${11/30}$ ${-11}/{(x+4)(x-7)}$=${11/30}$ $-30= (x^2-7x+4x-28)$ $x^2-3x+2=0$ Solving the eqn using quadratic formula, $x={-b±√{b^2-4ac}}/{2a}$ $x={3±√{9-8}}/{2}$ $x={3±1}/{2}$ i.e root tof the equations are $x=1$ and $x=2$4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is $1/3$. Find his present age.Solution Let Rehman's present age be $x$ Rehman's age 3 years ago = $x-3$ Rehman's age 5 years from now = $x+5$ Given: The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is $1/3$ i.e $1/{x-3} + 1/{x+5}=1/3$ ${(x+5)+(x-3)}/{(x-3)(x+5)} = 1/3$ $3(2x+2)=x^2+5x-3x-15$ $6x+6=x^2+2x-15$ $x^2-4x-21=0$ $x^2-7x+3x-21=0$ $x(x-7)+3(x-7)$ $(x-7)(x+3)$ So, Rehman's present age is 7 years.5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.Solution Let Shefali’s marks in Mathematics be x; and Shefali’s marks in English be (30-x); product of 2 marks more in Mathematics and 3 marks less in English = 210 Given: $(x+2)(30-x-3) = 210 $ $(x+2)(27-x) = 210$ $ 27x-x^2 +54 - 2x=210 $ $ x^2-25x+156=0 $ $x^2-13x-12x+156=0$ $x(x-13)-12(x-13)=0$ $(x-13)(x-12)=0$ So, the marks in Mathematics is 12 or 13 and the marks in English is 18 or 17.6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.Solution Let the shorter side be $x$ $m$; Therefore, longer side= $x+30$ $m$ and diagonal of the rectangle = $x+60$ $m$ Therefore $(x+60)^2=x^2 + (x+30)^2 $ $x^2+120x+3600 = x^2+x^2+60x+900$ $x^2-60x-2700=0$ Solving the eqn using quadratic formula, $x={-b±√{b^2-4ac}}/{2a}$ $x={60±√{3600-4(1)(-2700)}}/{2}$ $x={60±√14400}/{2}$ $x={60±120}/{2}$ $x=90$ Therefore, Breadth of the rectangle= 90 meters, length= 120 meters, and diagonal= 150 meters.$x+60$$x$$x+30$7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.Solution Let the two numbers be $x$ and $y$ where $x>y$ Therefore ,given, $x^2-y^2=180$ and $y^2=8x$ Therefore, $x^2-8x=180$ , $x^2-8x-180 =0$ Solving the eqn using quadratic formula, $x={-b±√{b^2-4ac}}/{2a}$ $x={8±√{64-4(1)(-180)}}/{2}$ $x={8±√784}/{2}$ $x={8±28}/{2}$ $x=18$ And $y^2=8x$ => $y^2=8(18)$ =>$y=√144$ $y=12$ Ans: The two number are 12 and 188. A train travels 360 km at a uniform speed. If the speed had been 5 km/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 $x$ km/h Therefore time taken =$t_1$= $360/x$ Now speed of the train in 2nd case = $x+5$ km/h Time taken =$t_2$= $t_1 - 1$ = $360/{x+5}$ => $360/x - 1 = 360/{x+5} $ => $360/x - 360/{x+5} = 1 $ => $360({1/x- 1/(x+5)}) = 1 $ => $360(x+5-x) = x(x+5) $ => $1800=x^2+5x$ => $x^2+5x -1800 = 0$Method 1: Solving by factorisation method, $x^2+5x -1800=0$ Splitting the middle term $5x=45x-40x$ $x^2+45x-40x -1800=0$ $x(x+45)-40(x+45)=0$ $(x+45)(x-40)=0$ Therefore x=40 km/h Ans: Speeed of the train= 40 km/hMethod 2 : Solving by quadratic formula. $x={-b±√{b^2-4ac}}/{2a}$ $x={-5±√{25+7200}}/{2}$ $x={-5±√7225}/{2}$ $x={-5±√(5× 5×17×17)}/{2}$ $x={-5±(5×17)}/{2}$ $x={-5±85}/{2}$ Therefore the roots of the quadratic equations are; $x= 80/2$ and $x= -90/2$ Since the speed of the train cannot be negative, hence the speed= $40$ km/h9. Two water taps together can fill a tank in 9$3/8$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.Solution Let capacity of tank be $V$ litres. and volume= flow × time Let flow of larger tap be =$Q_1$ and time taken to fill the tank be =$t_1$ =$x$ Hence $Q_1 = V/t_1$ Let flow of smaller tap be =$Q_2$ and time taken to fill the tank be =$t_2$ Hence $Q_2 = V/t_2$ Given : $t_2=t_1 - 10$ $t_2=x - 10$ When both the taps are open then the flow of water in tank is: $Q=Q_1+Q_2$ and time taken to fill the tank is $t= 75/8$ hr. $V/{75/8}=V/x + V/{x-10}$ $8/{75}=1/x + 1/{x-10}$ $8x(x-10)=75(x-10+x)$ $8x^2-80x-150x+750=0$ $8x^2-230x+750=0$ $4x^2-115x+375=0$ $4x^2-100x-15x+375=0$ $4x(x-25)-15(x-25)=0$ $(x-25)(4x-15)=0$ Therefore $x=25$ and $15/40$ Now larger diameter takes 10 hours less than smaller tap. So, $x-10$= 15 and $15/40-10$=> $-375/40$ Since the time taken cannot be negative, hence the time taken for smaller and larger taps are $25$ and $15$ hours respectively.Note: try to solve the equation using quadratic formula also 10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.Solution Let the speed of passanger train be $x$ km/h Therefore given,speed of express train = $x+11$ km/h Time taken by passanger train, $t_1$= $132/x$ Time taken by express train, $t_2$= $132/{x+11}$ Since, $t_2$= $t_1 - 1 $ Therefore: $(t_1 - 1)$= $132/{x+11}$ $(132/x - 1)$= $132/{x+11}$ $132/x$- $132/{x+11}$ = 1 ${132(x+11-x)}/{x(x+11)}$ = 1; $ 132×11=x^2+11x$; $ x^2+11x -1452 = 0$; Using quadratic formula; $x={-b±√{b^2-4ac}}/{2a}$ $x={-11±√{121+4(1452)}}/{2}$ $x={-11±√{5929}}/{2}$ $x={-11±77}/{2}$ SO the roots of the equations are $66/2$ and $-88/2$ Since the speed of train cannot be negative, therefore, speed of passanger train = 33 km/h therefore, speed of express train = 44 km/h11. Sum of the areas of two squares is 468 $m^2$. If the difference of their perimeters is 24 m, find the sides of the two squares.Solution Let the sides of the 1st square be $x$ m and the sides of the 2nd square be $y$ m Given:the difference of their perimeters is 24 m Therefore $4x-4y=24$ i.e $x-y=6$ ....................(1) and Sum of the areas of two squares is 468 $m^2$ i.e $x^2-y^2=468$ ............(2) Substuting $y$ in eqn (2) $x^2-(x-6)^2=468$ $x^2+x^2+36-12x=468$ $2x^2-12x-432=0$ $x^2-6x-216=0$ $x^2-18x+12x-216=0$ $x(x-18)+12(x-18)=0$ $(x-18)(x+12)=0$ Therefore $x=18$ m. Therefore sides of the square are 18 m and 12 m.