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Class 10 NCERT Solutions- Chapter 4 Quadratic Equations – Exercise 4.4
  • Last Updated : 03 Mar, 2021

Question 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

Solution:

(i) Given: 2x2-3x+5=0



Here a=2,b=-3 and c=5

\therefore Discriminant, D=b2-4ac

= (-3)2– 4 × 2 × 5)

= 9-40 = -31 < 0

Hence, the roots are imaginary.

(ii) Given: 3x2-4√3x + 4 = 0

Here a=3,b=√3 and c=4

\therefore Discriminant, D=b2-4ac



= (-4√3)2 – (4 × 3 × 4)

= 48 – 48 = 0

Hence, the roots are real and equal.

Using the formula,

x=\frac {-b\pm\sqrt{b^2-4ac}} {2a}       , we get 

x=\frac {-(-4\sqrt3)\pm\sqrt{(-4\sqrt3)^2-4\times3\times4}} {2\times3}

= \frac {4\sqrt3\pm\sqrt{48-48}} {6}=\frac {4\sqrt3} {6}=\frac {2} {\sqrt3}

Hence, the equal roots are \frac 2 {\sqrt3} and \frac 2 {\sqrt3}       .

(iii) Given: 2x2-6x+3=0

Here, a=2,b=-6 and c=3



\therefore Discriminant, D=b2-4ac

= (-6)2 – (4 × 2 × 3)

= 36 – 24 = 12 > 0

Hence, the roots are distinct and real.

Using the formula,

x=\frac {-b\pm\sqrt{b^2-4ac}} {2a}       ,we get

x=\frac {-(-6)\pm\sqrt{(-6)^2-4\times2\times3}} {2\times2}

x=\frac {6\pm\sqrt{36-24}} {4}

x=\frac {6\pm\sqrt{12}} {4}

x=\frac {6\pm2\sqrt{3}} {4}

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

Hence, the equal roots are \frac {3+\sqrt{3}} {2}and \frac {3-\sqrt{3}} {2}

Question 2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.

(i) 2x2+kx+3

(ii) kx(x-2)+6=0

Solution:

(i) 2x+kx+3=0

This equation is of the form ax2+bx+x, where a=2, b=k and c=3.

Discriminant, D=b2-4ac

=k – 4 × 2 × 3

=k2 -24

For equal roots D=0

\implies k2-24=0

\implies k2=24

\implies k2 = ±24 = ±2√6
 

(ii) kx(x-2)+6=0

\implies  kx2-2kx+6=0

This equation is of the form ax2+bx+c=0, where a=k, b=-2k and c=6.

Discriminant, D=b2-4ac

=(-2k)2 – 4 × k × 6

=4k2-24k

For equal roots D=0

\implies  4k2-24k=0

\implies  4k(k-24)=0

\implies k=0 (not possible) or 4k-24=0

\implies  k= 24/4=6

Question 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.

Solution:

Let the breadth of the rectangular mango grove be x m.

Then, the length of the rectangular mango grove will be 2x m.

The Area of the rectangular mango grove=length × breadth

According to the question, we have

x × 2x= 800

\implies  2x2=800

\implies  x2=400

\implies  x=20

Hence, the rectangular mango grove is possible to design whose length=40 m and breadth=20 m.

Question 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.

Solution:

Let the present age of one friend be x years.

Then, the present age of other friend be (20-x) years.

4 years ago, one friend’s age was (x-4) years

4 years ago, other friend’s age was (20-x-4)=(16-x) years.

According to the question,

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

\implies 16x-64-x2+4x=48

\implies x2-20x+112=0

This equation is of the form ax2+bx+c=0,where a=1, b=-20 and c=112.

Discriminant, D=b2-4ac

= (-20)2-4 × 1 × 112 = -48 < 0

Since, there are no real roots.

So the given situation is not possible.

Question 5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.

Solution:

Let the length of the rectangular park be x.

The perimeter of the rectangular park= 2(length + breadth)

\implies  2(x + breadth)=80

\implies  breadth=40-x

The area of rectangular park= length × breadth

\implies      x(40-x)=400

\implies 40x-x2=400

\implies x2-40x+400=0

\implies x2 -20x-20x+400=0

\implies  (x-20)(x-20)=0

\implies x=20

Hence, the rectangular park is possible to design. So, the length of the park is 20m and the breadth = 40-20=20m.

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