Class 10 NCERT Solutions Chapter 4 Quadratic Equations Exercise 4.4 (original) (raw)
Last Updated : 12 Aug, 2024
**Question 1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
****(i) 2x** 2 -3x+5=0
****(ii) 3x** 2 -4√3x+4=0
****(iii) 2x** 2 -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)** **2x 2 +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
=k2 - 4 × 2 × 3
=k2 -24
For equal roots D=0
\implies k2-24=0
\implies k2=24
\implies __k_2 = ±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 m 2 ? 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 m 2 ? 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.
Practice Questions - Quadratic Equations
**1).Find the nature of roots of the equation 2x² + x + 4 = 0.
**2).For what values of k will the equation kx² + 4x + 4 = 0 have real and equal roots?
**3).If α and β are the roots of the equation x² - px + q = 0, find the value of α² + β² in terms of p and q.
**4).Form a quadratic equation whose roots are 3 and -2.
**5).Without solving, determine if the following equation has real roots: x² + 2x + 5 = 0.
**6).Find the value of k for which the equation x² + kx + 64 = 0 has real roots.
**7).If the equation (1 + m²)x² + 2mcx + (c² - a²) = 0 has equal roots, prove that c² = a²(1 + m²).
**8).For what value of k will the roots of the equation x² - 2kx + k = 0 be reciprocals of each other?
**9).If the roots of ax² + bx + c = 0 are in the ratio p:q, show that b² : ac = (p + q)² : pq.
**10).Find the condition for the roots of ax² + bx + c = 0 to be irrational.
Summary:
Exercise 4.4 of Chapter 4 "Quadratic Equations" in Class 10 NCERT focuses on the nature of roots of quadratic equations. It explores the relationship between the discriminant (b² - 4ac) and the nature of roots (real and distinct, real and equal, or no real roots). Students learn to determine the nature of roots without solving the equation, find conditions for specific types of roots, and analyze the relationship between coefficients and roots. This exercise also covers forming quadratic equations when given information about their roots, and understanding the graphical representation of quadratic equations in relation to the nature of their roots.