JK Academy Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations Practice Set 2.1
Activity 1: (Textbook p. no. 31)
x2 + 3x – 5, 3x2 – 5x, 5x2; Write the polynomials In the index form.
Observe the coefficients and fill in the boxes.
Answer:
Index form of the given polynomials:
x2 + 3x – 5, 3x2 – 5x + 0, 5x2 + 0x + 0
i. Coefficients of x2 are [1], [3] and [5] respectively, and these coefficients are non zero.
ii. Coefficients of x are 3, [-5] and [0] respectively.
iii. Constant terms are [-5], [0] and [0] respectively.
Here constant terms of second and third polynomial is zero.
Activity 2: (Textbook p. no. 31)
Complete the following table
Answer:
| Quadratic Equation | General form | a | b | c |
| x2 – 4 = 0 | x2 +0x -4 =0 | 1 | 0 | -4 |
| y2 = 2y = 0 | y2 -2y + 7 = 0 | 1 | -2 | 7 |
| x2 + 2x = 0 | x2 + 2x + 0 = 0 | 1 | 2 | 0 |
Activity 3: (Textbook p. no. 32)
Decide which of the following are quadratic equations?
i. 9y2 + 5 = 0
Solution:
In the equation 9y2 + 5 = 0, y is the only variable and maximum index of the variable is 2.
∴ It is a quadratic equation.
ii. m3 – 5m2 + 4 = 0
Solution:
In the equation m3 – 5m2 + 4 = 0, m is the only variable and maximum index of the variable is not 2.
∴ It is not a quadratic equation.
iii. (l + 2)(l – 5) = 0
Solution:
(l + 2)(l – 5) = 0
∴ l(l – 5) + 2(l – 5) = 0
∴ l2 – 5l + 2l – 10 = 0
∴ l2 – 3l – 10 = 0.
In this equation l is the only variable and maximum index of the variable is 2
∴ it [is] a quadratic equation.
Activity 4: (Textbook p. no. 33)
If x = 5 is a root of equation kx2 – 14x – 5 = 0, then find the value of k by completing the following activity. Solution:
One of the roots of equation kx2 – 14x – 5 = 0 is 5
∴ Put x = 5 in the given equation.
∴ k52 – 14 (5) – 5 = 0
∴ 25k – 70 – 5 = 0
∴ 25k – 75 = 0
∴ 25k = 75
∴ k = 75/25 = 3
Practice Set 2.1
1. Write any two quadratic equations.
Solution:
i. m2 – 8y + 20 = 0
ii. a2 – 5a = 25
2. Decide which of the following are quadratic
i. x2 – 7y + 2 = 0
ii. y2 = 5y – 10
iii. y2 + 1/y = 2
iv. x + 1/x = -2
v. (m + 2) (m – 5) = 03
vi. m3 + 3m2 – 2 = 3m3
Solution:
i. The given equation is x2 + 5x – 2 = 0
Here, x is the only variable and maximum index of the variable is 2.
a = 1, b = 5, c = -2 are real numbers and a ≠ 0.
∴ The given equation is a quadratic equation.
iii. The given equation is
y2 + 1/y = 2
∴ y3 + 1 = 2y …[Multiplying both sides by y]
∴ y3 – 2y + 1 = 0
Here, y is the only variable and maximum index of the variable is not 2.
∴ The given equation is not a quadratic equation.
v. The given equation is
(m + 2) (m – 5) = 0
∴ m(m – 5) + 2(m – 5) = 0
∴ m2 – 5m + 2m – 10 = 0
∴ m2 – 3m – 10 = 0
Here, m is the only variable and maximum index of the variable is 2.
a = 1, b = -3, c = -10 are real numbers and a ≠ 0.
∴ The given equation is a quadratic equation.
vi. The given equation is
m3 + 3m2 – 2 = 3m3
∴ 3m3 – m3 – 3m2 + 2 = 0
∴ 2m3 – 3m2 + 2 = 0
Here, m is the only variable and maximum
index of the variable is not 2.
∴ The given equation is not a quadratic equation.
3. Write the following equations in the form ax2 + bx + c = 0, then write the values of a, b, c for each equation.
i. 2y = 10 – y2
ii. (x – 1)2 = 2x + 3
iii. x2 + 5x = – (3 – x)
iv. 3m2 = 2m2 – 9
v. P (3 + 6p) = – 5
vi. x2 – 9 = 13
Solution:
i. 2y – 10 – y2
∴ y2 + 2y – 10 = 0
Comparing the above equation with
ay2 + by + c = 0, we get
a = 1, b = 2, c = -10
ii. (x – 1)2 = 2x + 3
∴ x2 – 2x + 12x + 3
x2 – 2x + 1 – 2x – 30
∴ x2 – 4x – 2 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = 1, b = -4, c = -2
iii. x2 + 5x = – (3 – x)
∴ x2 + 5x = -3 + x
∴ x2 + 5x – x + 3 = 0
∴ x2 + 4x + 3 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = 1, b = 4, c = 3
iv. 3m2 = 2m2 – 9
∴ 3m2 – 2m2 + 9 = 0
∴ m2 + 9 = 0
∴ m2 + 0m + 9 = 0
Comparing the above equation with
am2 + bm + c = 0, we get
a = 1, b = 0, c = 9
v. p (3 + 6p) = – 5
∴ 3p + 6p2 = -5
∴ 6p2 + 3p + 5 = 0
Comparing the above equation with
ap2 + bp + c = 0, we get
a = 6, b = 3, c = 5
vi. x2 – 9 = 13
∴ x2 – 9 – 13 = 0
∴ x2 – 22 = 0
∴ x2 + 0x – 22 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = 1, b = 0, c = -22
4. Determine whether the values given against each of the quadratic equation are the roots of the equation.
i. x2 + 4x – 5 = 0; x = 1,-1
ii. 2m2 – 5m = 0; m = 2, 5/2
Solution:
i. The given equation is
x2 + 4x – 5 = 0 …(i)
Putting x = 1 in L.H.S. of equation (i), we get
L.H.S. = (1)2 + 4(1) – 5 = 1 + 4 – 5 = 0
∴ L.H.S. = R.H.S.
∴ x = 1 is the root of the given quadratic equation.
Putting x = -1 in L.H.S. of equation (i), we get
L.H.S. = (-1)2 + 4(-1) – 5 = 1 – 4 – 5 = -8
∴ LH.S. ≠ R.H.S.
∴ x = -1 ¡s not the root of the given quadratic equation.
ii. The given equation is
2m2 – 5m = 0 …(i)
Putting m = 2 in L.H.S. of equation (i), we get
L.H.S. = 2(2)2 – 5(2) = 2(4) -10 = 8 – 10 = -2
∴ L.H.S. ≠ R.H.S.
∴ m = 2 is not the root of the given quadratic equation.
Putting m = 5/2 in L.H.S. of equation (i), we get
L.H.S. = 2 (5/2)2 – 5 ( 5/2)
= 2 (25/4) – 25/2
= 25/2 – 25/2
= 0
∴ L.H.S. = R.H.S.
∴ m = 5/2 is the root of the given quadratic equation.
5. Find k if x = 3 is a root of equation kx2 – 10x + 3 = 0.
Solution:
x = 3 is the root of the equation kx2 – 10x + 3 = 0.
Putting x = 3 in the given equation, we get
k(3)2 – 10(3) + 3 = 0
∴ 9k – 30 +3 = 0
∴ 9k – 27 = 0
∴ 9k = 27
∴ k = 27/ 9
∴ k = 3
6. One of the roots of equation 5m2 + 2m + k = 0 is −7/5 Complete the following activity to find the value of ‘k’.
Solution:
– 7/5 is a root of quadratic equation
5m2 + 2m + k = 0
∴ put m= -7/5 in the equation
∴ 5 x ( -7/5)2 + 2 x -7/5 + k = 0
∴ 49/5 – 14/5 K = 0
∴ 7 + k = 0
∴ k = -7