10th Maths 4.3

Like, share, and Subscribe.

Exercise 4.3

Q1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square :

i) $2 x^{2} - 7 x + 3 = 0$ $2x^2-7x+3=0$

ii) $2 x^{2} + x - 4 = 0$ $2x^2+x-4=0$

iii) $4 x^{2} + 4 \sqrt{3} x + 3 = 0$ $4x^2+4sqrt3x+3=0$

iv) $2 x^{2} + x + 4 = 0$ $2x^2+x+4=0$

Sol. :

2 x^{2} - 7 x + 3 = 0

$2x^2-7x+3=0$ compare with

a x^{2} + b x + c = 0

$ax^2+bx+c=0$

where

a = 2, b = - 7

$a=2, b=-7$ and

c = 3

$c=3$

= b^{2} - 4 a c

$=b^2-4ac$

= {(- 7)}^{2} - 4 (2) (3)

$=(-7)^2-4(2)(3)$

= 49 - 24

$=49-24$

= 25 > 0

$=25>0$

The roots exist.

By completing the square

2 x^{2} - 7 x + 3 = 0

$2x^2-7x+3=0$

x^{2} - \frac{7 x}{2} + \frac{3}{2} = 0

$x^2-(7x)/2+3/2=0$

x^{2} - \frac{7 x}{2} = - \frac{3}{2}

$x^2-(7x)/2=-3/2$ [Add

{(\frac{7}{4})}^{2}

$(7/4)^2$ both sides]

x^{2} - \frac{7 x}{2} + {(\frac{7}{4})}^{2} = - \frac{3}{2} + {(\frac{7}{4})}^{2}

$x^2-(7x)/2+(7/4)^2=-3/2+(7/4)^2$

x^{2} - \frac{2 (7 x)}{2 (2)} + {(\frac{7}{4})}^{2} = - \frac{3}{2} + \frac{49}{16}

$x^2-(2(7x))/(2(2))+(7/4)^2=-3/2+49/16$

{(x - \frac{7}{4})}^{2} = \frac{- 24 + 49}{16}

$(x-7/4)^2=(-24+49)/16$

{(x - \frac{7}{4})}^{2} = \frac{25}{16}

$(x-7/4)^2=25/16$

{(x - \frac{7}{4})}^{2} = {(\frac{5}{4})}^{2}

$(x-7/4)^2=(5/4)^2$

{(x - \frac{7}{4})}^{2} - {(\frac{5}{4})}^{2} = 0

$(x-7/4)^2-(5/4)^2=0$ [

a^{2} - b^{2} = (a + b) (a - b)

$a^2-b^2=(a+b)(a-b)$ ]

(x - \frac{7}{4} - \frac{5}{4}) (x - \frac{7}{4} + \frac{5}{4}) = 0

$(x-7/4-5/4)(x-7/4+5/4)=0$

(x - \frac{12}{4}) (x - \frac{2}{4}) = 0

$(x-12/4)(x-2/4)=0$

(x - 3) (x - \frac{1}{2}) = 0

$(x-3)(x-1/2)=0$

x - 3 = 0

$x-3=0$

x = 3

$x=3$

x - \frac{1}{2} = 0

$x-1/2=0$

x = \frac{1}{2}

$x=1/2$

Answer :

x = 3, \frac{1}{2}

$x=3, 1/2$

ii)

2 x^{2} + x - 4 = 0

$2x^2+x-4=0$ compare with

a x^{2} + b x + c = 0

$ax^2+bx+c=0$

where

a = 2, b = 1

$a=2, b=1$ and

c = - 4

$c=-4$

= b^{2} - 4 a c

$=b^2-4ac$

= {(1)}^{2} - 4 (2) (- 4)

$=(1)^2-4(2)(-4)$

= 1 + 32

$=1+32$

= 33 > 0

$=33>0$

The roots exist.

By completing the square

2 x^{2} + x - 4 = 0

$2x^2+x-4=0$

x^{2} + \frac{x}{2} - \frac{4}{2} = 0

$x^2+x/2-4/2=0$

x^{2} + \frac{x}{2} = 2

$x^2+x/2=2$ [Add

{(\frac{1}{4})}^{2}

$(1/4)^2$ both sides]

x^{2} + \frac{x}{2} + {(\frac{1}{4})}^{2} = 2 + {(\frac{1}{4})}^{2}

$x^2+x/2+(1/4)^2=2+(1/4)^2$

{(x + \frac{1}{4})}^{2} = 2 + \frac{1}{16}

$(x+1/4)^2=2+1/16$

{(x + \frac{1}{4})}^{2} = \frac{32 + 1}{16}

$(x+1/4)^2=(32+1)/16$

{(x + \frac{1}{4})}^{2} = \frac{33}{16}

$(x+1/4)^2=33/16$

{(x + \frac{1}{4})}^{2} - {(\frac{\sqrt{33}}{4})}^{2} = 0

$(x+1/4)^2-((sqrt(33))/4)^2=0$ [

a^{2} - b^{2} = (a + b) (a - b)

$a^2-b^2=(a+b)(a-b)$ ]

(x + \frac{1}{4} + \frac{\sqrt{33}}{4}) (x + \frac{1}{4} - \frac{\sqrt{33}}{4}) = 0

$(x+1/4+sqrt(33)/4)(x+1/4-sqrt(33)/4)=0$

(x + \frac{1 + \sqrt{33}}{4}) = 0

$(x+(1+sqrt(33))/4)=0$

x = - \frac{1 + \sqrt{33}}{4}

$x=-(1+sqrt(33))/4$

(x + \frac{1 - \sqrt{33}}{4}) = 0

$(x+(1-sqrt(33))/4)=0$

x = - \frac{1 - \sqrt{33}}{4}

$x=-(1-sqrt(33))/4$

Answer :

x = - \frac{1 + \sqrt{33}}{4}, - \frac{1 - \sqrt{33}}{4}

$x=-(1+sqrt(33))/4, -(1-sqrt(33))/4$

iii)

4 x^{2} + 4 \sqrt{3} x + 3 = 0

$4x^2+4sqrt3x+3=0$ compare with

a x^{2} + b x + c = 0

$ax^2+bx+c=0$

where

a = 4, b = 4 \sqrt{3}

$a=4, b=4sqrt3$ and

c = 3

$c=3$

= b^{2} - 4 a c

$=b^2-4ac$

= {(4 \sqrt{3})}^{2} - 4 (4) (3)

$=(4sqrt3)^2-4(4)(3)$

= 48 - 48

$=48-48$

= 0

$=0$

The roots exist.

By completing the square

4 x^{2} + 4 \sqrt{3} x + 3 = 0

$4x^2+4sqrt3x+3=0$

{(2 x)}^{2} + 2 (2 x) (\sqrt{3}) + {(\sqrt{3})}^{2} = 0

$(2x)^2+2(2x)(sqrt3)+(sqrt3)^2=0$

{(2 x - \sqrt{3})}^{2} = 0

$(2x-sqrt3)^2=0$

2 x - \sqrt{3} = 0

$2x-sqrt3=0$

x = \frac{\sqrt{3}}{2}

$x=sqrt3/2$

Answer :

each

x = \frac{\sqrt{3}}{2}

$x=sqrt3/2$

iv)

2 x^{2} + x + 4 = 0

$2x^2+x+4=0$

x^{2} + \frac{x}{2} + 2 = 0

$x^2+x/2+2=0$ compare with

a x^{2} + b x + c = 0

$ax^2+bx+c=0$

where

a = 1, b = \frac{1}{2}

$a=1, b=1/2$ and

c = 2

$c=2$

= b^{2} - 4 a c

$=b^2-4ac$

= {(\frac{1}{2})}^{2} - 4 (1) (2)

$=(1/2)^2-4(1)(2)$

= \frac{1}{4} - 8

$=1/4-8$

= \frac{1 - 32}{4}

$=(1-32)/4$

= - \frac{31}{4} < 0

$=-31/4<0$

The root not exsits.

Answer :

Do not exist.

Q2. Find the roots of the quadratic equations by applying the quadratic formula.

i) $2 x^{2} - 7 x + 3 = 0$ $2x^2-7x+3=0$

ii) $2 x^{2} + x - 4 = 0$ $2x^2+x-4=0$

iii) $4 x^{2} + 4 \sqrt{3} x + 3 = 0$ $4x^2+4sqrt3x+3=0$

iv) $2 x^{2} + x + 4 = 0$ $2x^2+x+4=0$

Sol. :

2 x^{2} - 7 x + 3 = 0

$2x^2-7x+3=0$ compare with

a x^{2} + b x + c = 0

$ax^2+bx+c=0$

where

a = 2, b = - 7

$a=2, b=-7$ and

c = 3

$c=3$

Quadratic formula :

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

$x=(-b+-sqrt(b^2-4ac))/(2a)$

x = \frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 (2) (3)}}{2 \times 2}

$x=(-(-7)+-sqrt((-7)^2-4(2)(3)))/(2times2)$

x = \frac{7 \pm \sqrt{49 - 24}}{4}

$x=(7+-sqrt(49-24))/4$

x = \frac{7 \pm \sqrt{25}}{4}

$x=(7+-sqrt(25))/4$

We take (+)

$x=(7+5)/4$

$x=12/4$

$x=3$

We take (-)

$x=(7-5)/4$

$x=2/4$

$x=1/2$

Answer :

Roots are

$x=3, 1/2$ .

ii)

$2x^2+x-4=0$ compare with

$ax^2+bx+c=0$

where

$a=2, b=1$ and

$c=-4$

Quadratic formula :

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$x=(-1+-sqrt((1)^2-4(2)(-4)))/(2times2)$

$x=(-1+-sqrt(1+32))/4$

$x=(-1+-sqrt(33))/4$

We take (+)

$x=(-1+sqrt(33))/4$

We take (-)

$x=(-1-sqrt(33))/4$

$x=2/4=1/2$

Answer :

Roots are

$x=(-1+sqrt(33))/4, (-1-sqrt(33))/4$ .

iii)

$4x^2+4sqrt3x+3=0$ compare with

$ax^2+bx+c=0$

where

$a=4, b=4sqrt3$ and

$c=3$

Quadratic formula :

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$x=(-4sqrt3+-sqrt((4sqrt3)^2-4(4)(3)))/(2times4)$

$x=(-4sqrt3+-sqrt(48-48))/8$

$x=(-4sqrt3+-0)/8$

$x=(-4sqrt3)/8$

$x=-sqrt3/2$

Answer :

Roots are

$x=-sqrt3/2$ each.

iv)

$2x^2+x+4=0$ compare with

$ax^2+bx+c=0$

where

$a=2, b=1$ and

$c=4$

Quadratic formula :

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$x=(-1+-sqrt((1)^2-4(2)(4)))/(2times2)$

$x=(-1+-sqrt(1-32))/4$

$x=(-1+-sqrt(-31))/4$

Answer :

Do not exist.

Q3. Find the roots of the following equations :

i) $x-1/x=3, xne0$

ii) $1/(x+4)-1/(x-7)=11/30, xne-4, 7$

Sol. :

$x-1/x=3$

$x^2-1=3x$

$x^2-3x-1=0$ compare with

$ax^2+bx+c=0$

where

$a=1, b=-3$ and

$c=-1$

Quadratic formula :

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$x=(-(-3)+-sqrt((-3)^2-4(1)(-1)))/(2times1)$

$x=(3+-sqrt(9+4))/2$

$x=(3+-sqrt13)/2$

We take (+)

$x=(3+sqrt13)/2$

We take (-)

$x=(3-sqrt13)/2$

Answer :

$x=(3+sqrt13)/2, (3-sqrt13)/2$ .

ii)

$1/(x+4)-1/(x-7)=11/30, xne-4,7$

$(x-7-x-4)/((x+4)(x-7))=11/30$

$-11/(x^2-7x+4x-28)=11/30$

$11(x^2-3x-28)=-11(30)$

$x^2-3x-28+30=0$

$x^2-3x+2=0$

$x^2-2x-x+2=0$

$x(x-2)-1(x-2)=0$

$(x-2)(x-1)=0$

$x-2=0$

$x=2$

$x-1=0$

$x=1$

Answer :

$x=1, 2$ .

Q4. 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.

Sol. :

Let the present age of Rehman is

$x$ years,

$3$ years ago the age of Rehman is

$(x-3)$ years,

and

$5$ years from now the age of Rehman is

$(x+5)$ years.

According to question

$1/(x-3)+1/(x+5)=1/3$

$(x+5+x-3)/((x-3)(x+5))=1/3$

$(2x+2)/(x^2+5x-3x-15)=1/3$

$x^2+2x-15=6x+6$

$x^2+2x-6x-15-6=0$

$x^2-4x-21=0$

$x^2-7x+3x-21=0$

$x(x-7)+3(x-7)=0$

$(x-7)(x+3)=0$

$x-7=0$

$x=7$

$x+3=0$

$x=-3$ [ it is not possible]

Answer :

The present age of Rehman is

$7$ years.

Q5. 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.

Sol. :

Let the marks in Mathematics are

$x$ ,

then the marks in English are

$(30-x)$

According to question

$(x+2)times(30-x-3)=210$

$(x+2)times(27-x)=210$

$27x-x^2+54-2x=210$

$-x^2+25x+54-210=0$

$-x^2+25x-156=0$

$x^2-25x+156=0$

$x^2-12x-13x+156=0$

$x(x-12)-13(x-12)=0$

$(x-12)(x-13)=0$

$x-12=0$

$x=12$

$x-13=0$

$x=13$

when marks in Mathematics is

$12$ ,

then the marks in English

$30-12=18$ ,

when marks in Mathematics is

$13$ ,

then the marks in English is

$30-13=17$

Answer :

Marks in Mathematics is

$12$ , marks in English is

$18$ ,

Marks in Mathematics is

$13$ , marks in English is

$17$ .

Q6. 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.

Sol. :

Let the shroter side is

$x$ m,

diagonal is

$(x+60)$ m,

and longer side is

$(x+30)$ m.

According to question

$(x+60)^2=(x+30)^2+(x)^2$ [by Pythagoras therom ]

$x^2+120x+3600=x^2+60x+900+x^2$

$x^2+120x+3600-x^2-60x-900-x^2=0$

$-x^2+60x+2700=0$

$x^2-60x-2700=0$

$x^2-90x+30x-2700=0$

$x(x-90)+30(x-90)=0$

$(x-90)(x+30)=0$

$x-90=0$

$x=90$

$x+30=0$

$x=-30$ [ it is not possible]

shorter side is

$90$ m,

longer side is

$90+30=120$ m,

and diagonal is

$90+60=150$ m.

Answer :

shorter side

$90$ m, longer side

$120$ m and diagonal

$150$ m.

Q7. The difference of square of two numbers is $180$ . The square of the smaller number is $8$ times the larger number. Find the two numbers.

Sol. :

Let larger number is

$x$ ,

According to question

$(x)^2-8x=180$

$x^2-8x-180=0$

$x^2-18x+10x-180=0$

$x(x-18)+10(x-18)=0$

$(x-18)(x+10)=0$

$x-18=0$

$x=18$

$x+10=0$

$x=-10$ [it is not possible]

the larger number is

$18$

the smaller number is

$sqrt(8times18)=sqrt(144)=12$ .

Answer :

The larger number is

$18$ and the smaller is

$12$ .

Q8. 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.

Sol.

Let the first speed of the train is

$x$ km/h,

and the second speed of the train is

$x+5$ km/h,

travel time is distance/time

$=360/x hr$ and

$360/(x+5) hr$

According to question

$360/x-360/(x+5)=1$

$360(x+5)-360x=1$

$360x+1800-3600x=x(x+5)$

$1800=x^2+5x$

$x^2+5x-1800=0$

$x^2+45x-40x-1800=0$

$x(x+45)-40(x+45)=0$

$(x+45)(x-40)=0$

$x+45=0$

$x=-45$ [it is not possible]

$x-40=0$

$x=40$

Answer :

The speed of the train is

$40$ km/h.

Q9. 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.

Sol. :

Let smaller tap can fill in

$x$ hours,

then larger tap can fill in

$(x-10)$ hours,

and both tap can fill in

$9(3)/8=75/8$ hours.

According to question

$1/(x-10)+1/x=8/75$

$(x+x-10)/(x(x-10))=8/75$

$75(2x-10)=8(x(x-10))$

$150x-750=8x^2-80x$

$8x^2-80x-150x-750=0$

$8x^2-230x-750=0$

$8x^2-200x-30x-750=0$

$8x(x-25)-30(x-25)=0$

$(x-25)(8x-30)=0$

$x-25=0$

$x=25$

$8x-30=0$

$x=30/8=3.75$

Now

Time taken by larger tap

$=25-10=15$ hour,

Time taken by larger tap

$=3.75-10=-6.25$ hour [it is not possible]

Answer :

Smaller tap can fill

$25$ hr, and larger tap can fill

$15$ hr.

Q10. 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 station). 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.

Sol. :

Let the average speed of the passenger train is

$x$ km/h,

and the average speed of the express train is

$(x-10)$ km/h.

travel time distance/time

$=132/x hr$ and

$132/(x+11) hr$

According to question

$132/x-132/(x+11)=1$

$132(x+11)-132x=x(x+11)$

$132x+132times11-132x=x^2+11x$

$x^2+11x-132times11=0$

$x^2+44x-33x-132times11=0$

$x(x+44)-33(x+44)=0$

$(x+44)(x-33)=0$

$x+44=0$

$x=-44$ [it is not possible]

$x-33=0$

$x=33$ km/hr

the speed of express train

$=33+11=44$ km/hr.

Answer :

The average speed of express train

$44$ km/hr and average speed of passenger train

$33$ km/hr.

Q11. Sum of the areas of two squares is $468m^2$ . If the difference of their perimeters is $24m$ , find the sides of the two squares.

Sol. :

Side of suare is

$24/4=6$ m

Let the side of the first square is

$x$ m,

and the side of the second square is

$(x+6)$ m.

According to question

$(x+6)^2+x^2=468$

$x^2+12x+36+x^2=468$

$2x^2+12x+36-468=0$

$2x^2+12x-432=0$

$x^2+6x-216=0$

$x^2+18x-12x-216=0$

$x(x+18)-12(x+18)=0$

$(x+18)(x-12)=0$

$x+18=0$

$x=-18$ [ it is not possible]

$x-12=0$

$x=12$ m

The side of second square

$12+6=18$ m.

Answer :

The sides of square are

$12$ m,

$18$ m.

Search This Blog

10th Maths 4.3

NCERT Class 10th solution of Exercise 4.1

NCERT Class 10th solution of Exercise 4.2

NCERT Class 10th solution of Exercise 4.4

NCERT Class 10th Maths Projects

Exercise 4.3

Comments

Popular posts from this blog

10th Maths Chapter 1

CBSE 10th and 12th Result

RSKMP 5th & 8th Result