NCERT-class-10th - Quadratic equation - Exercise 4.4

Sum No.1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them.

(i) 2x²-3x+5=0 

(ii) 3x²-4√3x +4=0

(iii) 2x²-6x+3=0  

 

Ans.(i) 2x²-3x+5=0 

Comparing this equation with general equation ax²+bx+c=0,

We get a=2, b=-3 and c=5

Discriminant = b²-4ac=(-3)²-4*2*5 =9-40=-31

Discriminant is less than 0 which means equation has no real roots.

 

(ii) 3x²-4√3x +4=0

Comparing this equation with general equation ax²+bx+c=0,

We get a=3, b=-4√3 and c=4

Discriminant = b²-4ac=(-4√3)²-4*3*4=48-48=0

Discriminant is equal to 0 which means equation has equal real roots.

Applying quadratic  x= [-b 士√b²-4ac]/ 2a to find roots

⇒x= [4√3 士√0]/ 2*3 =(2√3) /3

Because, equation has two equal roots, it means 

x= (2√3)/3 , (2√3)/3

 

(iii) 2x²-6x+3=0 

Comparing this equation with general equation ax²+bx+c=0,

We get a=2, b=-6 and c=3

Discriminant=b²-4ac=(-6)²-4*2*3 =36-24=12

Value of discriminant is greater than zero.

Applying quadratic  x= [-b 士√b²-4ac]/ 2a  to find roots

⇒x= [6 士√12]/ 2*2= [6 士2√3]/ 4

⇒x=(3士√3)/2

⇒x=(3+√3)/2, (3-√3)/2

Sum No.2. Find the value of k for each of the following quadratic equations, so that they have two equal roots.

(i)2x²+kx+3=0

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

 

Ans. (i)2x²+kx+3=0

Comparing equation 2x²+kx+3=0 with general quadratic equation ax²+bx+c=0, we get a=2,b=k and c=3

Discriminant =b²-4ac =k²-4*2*3= k²-24

we know that quadratic equation has two equal roots only when the values of discriminant is equal to zero

Putting discriminant equal to zero

k²-24=0 ⇒ k²=24

⇒k =士 √24 =士2√6

⇒k =2√6,-2√6

 

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

⇒kx²-2kx+6=0

Comparing equation kx²-2kx+6=0 with general quadratic equation ax²+bx+c=0, we get a=k, b=-2k and c=6

Discriminant =b²-4ac =(-2k)²-4*k*6= 4k²-24k

we know that quadratic equation has two equal roots only when the values of discriminant is equal to zero

Putting discriminant equal to zero

4k²-24k=0 ⇒4k(k-6)=0

⇒k =0 , 6

The basic definition of quadratic equation says that quadratic equation is the equation of the form ax²+bx+c=0, where a is not equal to zero.

 Therefore,  in equation kx²-2kx+6=0, we cannot have k=0.

Therefore, we discard k=0.

Hence the answer is k=6.

 

Sum No.3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m². If so, find its length and breadth.

Ans. Let breadth of rectangular mango grove = x metres

Let lenght of rectangular mango grove = 2x metres

Area of rectangle = length*breadth=x*2x=2x²m²

Acc. to the given condition:

2x²=800 ⇒ 2x²-800=0⇒  x²-400=0

Comparing equation x²-400=0 with general form of quadratic equation ax²+bx+c=0, we get a=1,b=0 and c=-400

Discriminant = b²-4ac = (0)²-4(1)(-400)=1600

Discriminant is > 0 means that equation has two distinct real roots.

Therefore, it is possible to design a rectangular grove.

Applying quadratic formula, x= [-b 士√b²-4ac]/ 2a  to find roots

⇒x= [0 士√1600]/ 2*1=  士 40/2=士20

 ⇒ x=20,-20 

We discard negative value of x because breadth of rectangle cannot be in negative.

Therefore, x= breadth of rectangle =20 metres

Length of rectangle = 2x= 2*20= 40 metres

 

Sum No.4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Ans. Let age of first friend =x years

Let age of second friend = (20-x) years

Four year ago, age of first friend = (x-4) year

Four year ago, age of second friend= (20-x)-4=(16-x) years

Acc. to given condition,

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

⇒ 16x-x²-64+4x=48

⇒ 20x-x²-112=0 

⇒ x²-20x-112=0

Comparing equation, x²-20x+112=0 with general quadratic equation ax²+bx+c=0, we get a=1,b=-20 and c=112

Discriminant =b^2—4ac= (-20)²-4*1*112=400-448=-48<0 

Discriminant is less than zero which means we have no real roots for this equation.

Therefore, the given situation is not possible.

Sum No.5. Is it possible to design a rectangular park of perimeter 80 metres and area 400 m². If so, find its length and breadth.

Ans. Let length of park = x metres

We are given area of rectangular park= 400 m²

Now, breadth of the park = 400/x metres

Area of rectangle = length*breadth

Perimeter of rectangular park =2(length*breadth)= 2[x+(400/x)] metres

We are given perimeter of rectangle= 80 metres

Acc. to the condition:-

2[x+(400/x)]= 80

⇒2[(x²+400)/x]=80 

⇒2x²+800=80x 

⇒2x²-80x+800=0 

⇒ x²-40x+400=0 

Comparing equation x²-40x+400=0 with general quadratic equation ax²-bx+c=0,we get a=1, b=-40 and c=400

Discriminant =b²-4ac=(-40)²-4*1*400=1600-1600=0

Discriminant=0.

Therefore, two roots of equation have real and equal roots, which means that it is possible to design a reactangular park of perimeter 80 m and area 400 m². Applying quadratic  x= [-b 士√b²-4ac]/ 2a to find roots

⇒x= [40士√(-40)²-4*1*400]/ 2*1= [40士√0]/ 2*1

⇒x=40/2

⇒x=20

Here, both the roots are equal to 20

Here, both the roots are equal to 20.

Therefore, length of rectangular park = 20 metres

Breadth of rectangular park =400/x=400/20=20 m 

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