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NCERT Solutions for Class 10 Maths Exercise 2.2 (NEW SYLLABUS)
Solve the followings Questions.
1. Find the zeros of the quadratic polynomials and verify the relationships between the zeros and the coefficients.
(i) x2-2x-8
Answer:
Let, P(x) = x2-2x-8
Comparing P(x) with a quadratic equation ax2+bx+c
we have b = -2 and c = -8.
P(x) = x2-2x-8
= x2-4x+2x-8
= x(x-4) + 2(x-4)
= (x-4) (x+2)
Therefore, the value of P(x) will be zero if x-4=0 or x+2=0 i.e., when x=4 or x=-2
Hence the zeros of P(x) are 4, -2
Sum of zeros = 4+(-2) = 2/1 = –-b/a = (-Coefficient of x)/(Coefficient of x2)
Product of the zeros = 4(-2) = -8 = -8/1 = c/a = Constant term / Coefficient of x2
(ii) 4s2-4s+1
Answer:
Let P(x) = 4s2-4s+1
To find the zeros of the quadratic polynomial we consider
4s2 − 4s + 1 = 0
4s2 − 2s – 2s + 1 = 0
2s(2s – 1) -1(2s – 1) = 0
(2s – 1)(2s – 1) = 0
2s – 1 = 0 and 2s – 1 = 0
2s = 1 and 2s = 1
s = 1/2 and s = 1/2
∴ The zeroes of the polynomial = 1/2 and 1/2
Sum of the zeroes = -(coefficient of s)/(coefficient of s2)
Sum of the zeroes = -(-4)/4 = 1
Let’s find the sum of the roots = 1/2 + 1/2 = 1
Product of the zeros = Constant term / Coefficient of s2
Product of the zeros =1 / 4
Let’s find the products of the roots = 1/2 × 1/2 = 1/4
(iii) 6x2-3-7x
Answer:
Let, P(x) = 6x2-3-7x
To find the zeros
Let us put f(x) = 0
⇒ 6x2 – 7x – 3 = 0
⇒ 6x2 – 9x + 2x – 3 = 0
⇒ 3x(2x – 3) + 1(2x – 3) = 0
⇒ (2x – 3)(3x + 1) = 0
⇒ 2x – 3 = 0
x = 3/2
⇒ 3x + 1 = 0
⇒ x = -1/3
It gives us 2 zeros, for x = 3/2 and x = -1/3
Hence, the zeros of the quadratic equation are 3/2 and -1/3.
Now, for verification
Sum of zeros = – coefficient of x / coefficient of x2
3/2 + (-1/3) = – (-7) / 6 7/6 = 7/6
Product of roots = constant / coefficient of x2
3/2 x (-1/3) = (-3) / 6 -1/2 = -1/2
Therefore, the relationship between zeros and their coefficients is verified.
(iv) 4u2+8u
Answer:
Let, P(u) = 4u2+8u
Comparing P(u) with a quadratic equation au2+bu+c we have a = 4 ,b= 8 and c = 0
P(u) = 4u2+8u
= u(4u+8)
Therefore, the value of P(u) will be zero
Then, u = 0
Or, 4u+8 = 0
4u = -8
u = -2
Hence the zeros of P(s) are 0, -2
Now sum of zeros = 0+(-2) =-2 = -2
And product of the zeros = 0 X 2= 0
But, the Sum of the zeroes in any quadratic polynomial equation is given by = −coeff.of u / coeff.of u2 = −8/4 = −2
and, the product of zeroes in any quadratic polynomial equation is given by = constant term / coeff.of u2 = −0/4 = 0
(v) t2-15
Answer:
Let P(t) = t2-15
Comparing P(t) with a quadratic equation at2+bt+c we have a = 1,b= 0 and c = -15
Therefore, the value of P(t) will be zero if t2-15 = 0 i.e., t2= 15 so t = ±√15
Hence the zeros of P(s) are √15, – √15
Sum of zeros = √15 – √15 = 0 = (−coeff.of t / coeff.of t2)
Product of the zeros = √15(-√15) = -15 =(constant term / coeff.of t2)
(vi) 3x2– x-4
Answer:
Let P(x) = 3x2-x-4
Comparing P(x) with a quadratic equation ax2+bx+c we have a = 3, b= -1 and c = –4
P(x) = 3x2-x-4
= 3x2-4x+3x-4
= x(3x-4)+1(3x-4)
= (3x-4)(x+1)
Therefore the value of P(x) will be zero if 3x-4=0 or x+1=0
Then, 3x = 4
x = 4/3
Or, x = -1
Hence the zeros of P(x) are 4/3, – 1
Sum of zeros = 4/3 – 1= 1/3 = -b/a = (-Coefficient of x)/(Cofficient of x2)
Product of the zeros = (4/3) (-1) = –4/3 = = (constant term)/(Cofficient of x2)
2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 1/4 , −1
(ii) √2 , 13
(iii) 0, √5
(iv) 1, 1
(v) -1/4 , 1/4
(vi) 4, 1
Answer:
(i) 1/4, -1
Now formula of quadratic equation is
x²-(Sum of root)x + (Product of root) = 0
Plug the value in formula we get
x² –(1/4)x -1 = 0
Multiply by 4 to remove denominator we get
4x² – x -4 = 0
(ii) √2 , 1/3
Now formula of quadratic equation is
x²-(Sum of root)x + (Product of root) = 0
Plug the value in formula we get
x² –(√2)x + 1/3 = 0
Multiply by 3 to remove denominator we get
3x² – 3√2 x + 1 = 0
(iii) 0, √5
Now formula of quadratic equation is
x²-(Sum of root)x + (Product of root) = 0
Plug the value in formula we get
x² –(0)x + √5 = 0
simplify it we get
x² + √5 = 0
(iv) 1, 1
Now formula of quadratic equation is
x²-(Sum of root)x + (Product of root) = 0
Plug the value in formula we get
x² –(1)x + 1 = 0
simplify it we get
x² – x + 1 = 0
(v) -1/4 ,1/4
Now formula of quadratic equation is
x²-(Sum of root)x + (Product of root) = 0
Plug the value in formula we get
x² –(-1/4)x + 1/4 = 0
multiply by 4 we get
4x² + x + 1 = 0
(vi) 4, 1
Now formula of quadratic equation is
x²-(Sum of root)x + (Product of root) = 0
Plug the value in formula we get
x² –(4)x + 1 = 0
x² – 4x + 1 = 0
| CHAPTER NAME | OLD NCERT | NEW NCERT | |
| Real Numbers | EXERCISE 1.1 | ||
| EXERCISE 1.2 | 1.1 | CLICK HERE | |
| EXERCISE 1.3 | 1.2 | CLICK HERE | |
| EXERCISE 1.4 | |||
| Polynomials | EXERCISE 2.1 | 2.1 | CLICK HERE |
| EXERCISE 2.2 | 2.2 | CLICK HERE | |
| EXERCISE 2.3 | |||
| EXERCISE 2.4 | |||
| Pair of Linear Equations in Two Variables | EXERCISE 3.1 | ||
| EXERCISE 3.2 | 3.1 | CLICK HERE | |
| EXERCISE 3.3 | 3.2 | CLICK HERE | |
| EXERCISE 3.4 | 3.3 | CLICK HERE | |
| EXERCISE 3.5 | |||
| EXERCISE 3.6 | |||
| EXERCISE 3.7 | |||
| Quadratic Equations | EXERCISE 4.1 | 4.1 | CLICK HERE |
| EXERCISE 4.2 | 4.2 | CLICK HERE | |
| EXERCISE 4.3 | |||
| EXERCISE 4.4 | 4.3 | CLICK HERE | |
| Arithmetic Progressions | EXERCISE 5.1 | 5.1 | CLICK HERE |
| EXERCISE 5.2 | 5.2 | CLICK HERE | |
| EXERCISE 5.3 | 5.3 | CLICK HERE | |
| EXERCISE 5.4 | 5.4 (Optional) | CLICK HERE | |
| Triangles | EXERCISE 6.1 | 6.1 | CLICK HERE |
| EXERCISE 6.2 | 6.2 | CLICK HERE | |
| EXERCISE 6.3 | 6.3 | CLICK HERE | |
| EXERCISE 6.4 | |||
| EXERCISE 6.5 | |||
| EXERCISE 6.6 | |||
| Coordinate Geometry | EXERCISE 7.1 | 7.1 | CLICK HERE |
| EXERCISE 7.2 | 7.2 | CLICK HERE | |
| EXERCISE 7.3 | |||
| EXERCISE 7.4 | |||
| Introduction to Trigonometry | EXERCISE 8.1 | 8.1 | CLICK HERE |
| EXERCISE 8.2 | 8.2 | CLICK HERE | |
| EXERCISE 8.3 | |||
| EXERCISE 8.4 | 8.3 | CLICK HERE | |
| Some Applications of Trigonometry | EXERCISE 9.1 | 9.1 | CLICK HERE |
| Circles | EXERCISE 10.1 | 10.1 | CLICK HERE |
| EXERCISE 10.2 | 10.2 | CLICK HERE | |
| Construction | |||
| Areas Related to Circles | EXERCISE 12.1 | ||
| EXERCISE 12.2 | 11.1 | CLICK HERE | |
| EXERCISE 12.3 | |||
| Surface Areas and Volumes | EXERCISE 13.1 | 12.1 | CLICK HERE |
| EXERCISE 13.2 | 12.2 | CLICK HERE | |
| EXERCISE 13.3 | |||
| EXERCISE 13.4 | |||
| EXERCISE 13.5 | |||
| Statistics | EXERCISE 14.1 | 13.1 | CLICK HERE |
| EXERCISE 14.2 | 13.2 | CLICK HERE | |
| EXERCISE 14.3 | 13.3 | CLICK HERE | |
| EXERCISE 14.4 | |||
| Probability | EXERCISE 15.1 | 14.1 | CLICK HERE |
| EXERCISE 15.2 |