10th Maths 4.2

Exercise 4.2

Q1. Find the roots of the following quadratic equation by factorisation :

i) $x^2-3x-10=0,$

ii) $2x^2+x-6=0,$

iii) $sqrt2x^2+7x+5sqrt2=0,$

iv) $2x^2-x+1/8=0,$

v) $100x^2-20x+1=0.$

Sol. :

$x^2-3x-10=0$

$x^2-5x+2x-10=0$

$x(x-5)+2(x-5)=0$

$(x-5)(x+2)=0$

$x-5=0$

$x=5$

$x+2=0$

$x=-2$

Answer :

$x=5, x=-2$ .

ii)

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

$2x^2+4x-3x-6=0$

$2x(x+2)-3(x+2)=0$

$(x+2)(2x-3)=0$

$x+2=0$

$x=-2$ ,

$2x-3=0$

$x=3/2$

Answer :

$x=-2, x=3/2$ .

iii)

$sqrt2x^2+7x+5sqrt2=0$

$sqrt2x^2+5x+2x+5sqrt2=0$

$x(sqrt2x+5)+sqrt2(sqrt2x+5)=0$

$(sqrt2x+5)(x+sqrt2)=0$

$sqrt2x+5=0$

$x=-5/sqrt2$

$x+sqrt2=0$

$x=-sqrt2$

Answer :

$x=-5/sqrt2, x=-sqrt2$ .

iv)

$2x^2-x+1/8=0$

$16x^2-8x+1=0$

$16x^2-4x-4x+1=0$

$4x(4x-1)-1(4x-1)=0$

$(4x-1)(4x-1)=0$

$(4x-1)^2=0$

$4x-1=0$

$x=1/4$

Answer :

each

$x=1/4$ .

$100x^2-20x+1=0$

$100x^2-10x-10x+1=0$

$10x(10x-1)-1(10x-1)=0$

$(10x-1)(10x-1)=0$

$(10x-1)^2=0$

$10x-1=0$

$x=1/10$

Answer :

$x=1/10, x=1/10$ .

Q2. i) John and Jivanti together have $45$ marbles. Both of them lost $5$ marbles each and the product of the number of marbles they now have is $124.$ Find out how many marbles they had to start with.

ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in $₹$ ) was found to be $55$ minus the number of toys produced in a day. On a particular day, the total cost of production was $₹750$ . Find out the number of toys produced on that day.

Sol. :

Let the John had

$x$ marbles,

and the Jivanti had

$(45-x)$ marbles.

According to question

$(x-5)(45-x-5)=124$

$(x-5)(40-x)=124$

$x(40-x)-5(40-x)=124$

$40x-x^2-200+5x=124$

$-x^2+45x-200-124=0$

$x^2-45x-324=0$

$x^2-9x-36x-324=0$

$x(x-9)-36(x-9)=0$

$(x-9)(x-36)=0$

$x-9=0$

$x=9$

$x-36=0$

$x=36$

Answer :

John had

$9$ marbles and Jivanti had

$36$ marbles.

ii)

Let the number of toys produced on a day is

$x$ ,

and the cost of each toys

$₹(55-x)$

According to question

$xtimes(55-x)=750$

$55x-x^2=750$

$x^2-55x+750=0$

$x^2-25x-30x+750=0$

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

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

$x-25=0$

$x=25$

$x-30=0$

$x=30$

Answer :

The number of toys produced a day was

$25$ or

$30$ .

Q3. Find two numbers whoes sum is $27$ and product is 182`.

Sol. :

Let the first number is

$x$ ,

and the second number

$(27-x)$

According to question

$xtimes(27-x)=182$

$27x-x^2=182$

$x^2-27x+182=0$

$x^2-13x-14x+182=0$

$x(x-13)-14(x-13)=0$

$(x-13)(x-14)=0$

$x-13=0$

$x=13$

$x-14=0$

$x=14$

Answer :

Numbers are

$13$ and

$14$ .

Q4. Find two consecutive positive integers, sum of whose square is $365$ .

Sol.

Let the first number is

$x$ ,

and the second number is

$(x+1)$

According to question

$(x)^2+(x+1)^2=365$

$x^2+x^2+2x+1=365$

$2x^2+2x+1-365=0$

$2x^2+2x-364=0$

$x^2+x-182=0$

$x^2+14x-13x-182=0$

$x(x+14)-13(x+14)=0$

$(x+14)(x-13)=0$

$x+14=0$

$x=-14$

$x-13=0$

$x=13$

$x+1=13+1=14$

Answer :

Positive integers are

$13$ and

$14$ .

Q5. Altitude of a right triangle is $7$ cm less than its base. If the hypotenuse is $13$ cm, find the other two sides.

Sol. :

Let the base of a triangle is

$x$ cm,

the Altitude of a triangle is

$(x-7)$ cm, and

the hypotenuse of a triangle is

$13$ cm.

According to question

$(x)^2+(x-7)^2=(13)^2$

$x^2+x^2-14x+49=169$

$2x^2-14x+49-169=0$

$2x^2-14x-120=0$

$x^2-7x-60=0$

$x^2-12x+5x-60=0$

$x(x-12)+5(x-12)=0$

$(x-12)(x+5)=0$

$x-12=0$

$x=12$

$x+5=0$

$x=-5$

$x-7=12-7=5$

Answer :

The base is

$12$ cm and the altitude is

$5$ cm.

Q6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in $₹$ ) was $3$ more than twice the number of articles produced in that day was $₹90$ . Find the number of articles produced and the cost of each article.

Sol. :

Let the number of articles produced a days was

$x$ ,

and the cost of each article was

$₹(2x+3)$ ,

According to question

$xtimes(2x+3)=90$

$2x^2+3x=90$

$2x^2+3x-90=0$

$2x^2+15x-12x-90=0$

$x(2x+15)-6(2x+15)=0$

$(2x+15)(x-6)=0$

$2x+15=0$

$x=-15/2$ ( it is not possible)

$x-6=0$

$x=6$

$2x+3=2times6+3=15$

Answer :

The number of articles

$=6$ , cost of each article

$=₹15$ .

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10th Maths 4.2

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