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
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Exercise 4.3
Q1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square :
i) `2x^2-7x+3=0`
ii) `2x^2+x-4=0`
iii) `4x^2+4sqrt3x+3=0`
iv) `2x^2+x+4=0`
Sol. :
i)
`2x^2-7x+3=0` compare with `ax^2+bx+c=0`
where `a=2, b=-7` and `c=3`
`=b^2-4ac`
`=(-7)^2-4(2)(3)`
`=49-24`
`=25>0`
The roots exist.
By completing the square
`2x^2-7x+3=0`
`x^2-(7x)/2+3/2=0`
`x^2-(7x)/2=-3/2` [Add `(7/4)^2` both sides]
`x^2-(7x)/2+(7/4)^2=-3/2+(7/4)^2`
`x^2-(2(7x))/(2(2))+(7/4)^2=-3/2+49/16`
`(x-7/4)^2=(-24+49)/16`
`(x-7/4)^2=25/16`
`(x-7/4)^2=(5/4)^2`
`(x-7/4)^2-(5/4)^2=0` [`a^2-b^2=(a+b)(a-b)`]
`(x-7/4-5/4)(x-7/4+5/4)=0`
`(x-12/4)(x-2/4)=0`
`(x-3)(x-1/2)=0`
`x-3=0`
`x=3`
`x-1/2=0`
`x=1/2`
Answer :
`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`
`=b^2-4ac`
`=(1)^2-4(2)(-4)`
`=1+32`
`=33>0`
The roots exist.
By completing the square
`2x^2+x-4=0`
`x^2+x/2-4/2=0`
`x^2+x/2=2` [Add `(1/4)^2` both sides]
`x^2+x/2+(1/4)^2=2+(1/4)^2`
`(x+1/4)^2=2+1/16`
`(x+1/4)^2=(32+1)/16`
`(x+1/4)^2=33/16`
`(x+1/4)^2-((sqrt(33))/4)^2=0` [`a^2-b^2=(a+b)(a-b)`]
`(x+1/4+sqrt(33)/4)(x+1/4-sqrt(33)/4)=0`
`(x+(1+sqrt(33))/4)=0`
`x=-(1+sqrt(33))/4`
`(x+(1-sqrt(33))/4)=0`
`x=-(1-sqrt(33))/4`
Answer :
`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`
`=b^2-4ac`
`=(4sqrt3)^2-4(4)(3)`
`=48-48`
`=0`
The roots exist.
By completing the square
`4x^2+4sqrt3x+3=0`
`(2x)^2+2(2x)(sqrt3)+(sqrt3)^2=0`
`(2x-sqrt3)^2=0`
`2x-sqrt3=0`
`x=sqrt3/2`
Answer :
each `x=sqrt3/2`
iv)
`2x^2+x+4=0`
`x^2+x/2+2=0` compare with `ax^2+bx+c=0`
where `a=1, b=1/2` and `c=2`
`=b^2-4ac`
`=(1/2)^2-4(1)(2)`
`=1/4-8`
`=(1-32)/4`
`=-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) `2x^2-7x+3=0`
ii) `2x^2+x-4=0`
iii) `4x^2+4sqrt3x+3=0`
iv) `2x^2+x+4=0`
Sol. :
i)
`2x^2-7x+3=0` compare with `ax^2+bx+c=0`
where `a=2, b=-7` and `c=3`
Quadratic formula :
`x=(-b+-sqrt(b^2-4ac))/(2a)`
`x=(-(-7)+-sqrt((-7)^2-4(2)(3)))/(2times2)`
`x=(7+-sqrt(49-24))/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. :
i)
`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`,
or
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`,
or
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.
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