10th Maths 4.4

Exercise 4.4

Q1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them ;

i) $2 x^{2} - 3 x + 5 = 0$ $2x^2-3x+5=0$

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

iii) $2 x^{2} - 6 x + 3 = 0$ $2x^2-6x+3=0$

Sol. :

2 x^{2} - 3 x + 5 = 0

$2x^2-3x+5=0$ compare with

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

$ax^2+bx+c=0$

where

a = 2, b = - 3

$a=2, b=-3$ and

c = 5

$c=5$

The Discriminant

D = b^{2} - 4 a c

$D =b^2-4ac$

D = {(- 3)}^{2} - 4 (2) (5)

$D =(-3)^2-4(2)(5)$

D = 9 - 40

$D =9-40$

D = - 31 < 0

$D =-31<0$

no real roots.

Answer :

Real roots do not exist.

ii)

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

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

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

$ax^2+bx+c=0$

Where

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

$a=3, b=4sqrt3$ and

c = 4

$c=4$

The Discriminant

D = b^{2} - 4 a c

$D =b^2-4ac$

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

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

D = 48 - 48

$D =48-48$

D = 0

$D =0$

two equal roots.

Now

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

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

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

$x=(-(4sqrt3)+-sqrt(b^2-4ac))/(2times3)$

$x=(4sqrt3+-0)/6$

$x=(4sqrt3)/6$

$x=(2sqrt3)/3$

$x=(2sqrt3)/(sqrt3timessqrt3)$

$x=2/sqrt3$

Answer :

Equal roots ;

$2/sqrt3, 2/sqrt3$ .

iii)

$2x^2-6x+3=0$ compare with

$ax^2+bx+c=0$

Where

$a=2, b=-6$ , and

$c=3$

The Discriminant D

$D =b^2-4ac$

$D =(-6)^2-4(2)(3)$

$D =36-24$

$D =12>0$

unequal real root.

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

$x=(-(-6)+-sqrt(12))/(2(2))$

$x=(6+-sqrt(12))/4$

$x=(6+-2sqrt3)/4$

$x=(3+-sqrt3)/2$

Answer :

Unequal roots ;

$(3+sqrt3)/2, (3-sqrt3)/2$ .

Q2. Find the values of $k$ for each of the following quadratic equations, so that they have two equal roots.

i) $2x^2+kx+3=0$

ii) $kx(x-2)+6=0$

Sol. :

$2x^2+kx+3=0$ compare with

$ax^2+bx+c=0$

Where

$a=2,b=k$ , and

$c=3$

The roots are equal then

The Discriminant D

$=0$

$D =b^2-4ac$

$0 =k^2-4(2)(3)$

$0 =k^2-24$

$k^2 =24$

$k =+-2sqrt6$

Answer :

$k=+-2sqrt6$ .

ii)

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

$kx^2-2kx+6=0$ compare with

$ax^2+bx+c=0$

Where

$a=k, b=-2k$ and

$c=6$

The roots are equal then

The Discriminant D

$=0$

$D =b^2-4ac$

$0=(-2k)^2-4(k)(6)$

$0=4k^2-24k$

$0=4k(k-6)$

$0=k-6$

$k=6$

Answer :

$k=6$ .

Q3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is $800m^2$ ? If so, find its length and breadth.

Sol. :

Let the breadth of the mango grove is

$x$ m,

and the length of the mango grove is

$2x$ m,

According to question

$x(2x)=800$

$2x^2=800$

$x^2=400$

$x=+-20$

negative breadth is not possible, so

breadth is

$20$ m.

length is

$2(20)=40$ m

Answer :

Yes,

$40$ m,

$20$ m.

Q4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is $20$ years ago, the product of their ages in years was $48$ .

Sol.

Let the present age of the first friend is

$x$ years

and the present age of the second friend is

$(20-x)$ years.

According to question

$(x-4)(20-x-4)=48$

$(x-4)(16-x)=48$

$x(16-x)-4(16-x)=48$

$16x-x^2-64+4x=48$

$-x^2+20x-64-48=0$

$x^2-20x+112=0$ compare with

$ax^2+bx+c=0$

Where

$a=1, b=-20$ and

$c=112$

$D=b^2-4ac$

$D=(-20)^2-4(1)(112)$

$D=400-448$

$D=-48<0$ so, given situation not possible.

Answer :

No, given situation not possible.

Q5. Is it possible to design a rectangular park of perimeter $80m$ and area $400m^2$ ? If so, find its length and breadth.

Sol. :

Let the breadth of the park is

$x$ m.

and length of the park is

$((perimeter)/2-breadth)$

$=80/2-x=(40-x)$ m

According to question

$x(40-x)=400$

$40x-x^2=400$

$x^2-40x+400=0$ compare with

$ax^2+bx+c=0$

Where

$a=1$ , b=-40

$and$ c=400`

$D=b^2-4ac$

$D=(-40)^2-4(1)(400)$

$D=1600-1600$

$D=0$ so, roots are equal.

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

$x=(-(-40)+-sqrt(0))/(2(1))$

$x=40/2=20$ m

length

$40-20=20$ m

Answer :

Yes, length

$20$ m, breadth

$20$ m.

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