Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

Question 1.
Write any two quadratic equations.
Solution:
i. y² – 7y + 12 = 0
ii. x² – 8 = 0

Question 2.
Decide which of the following are quadratic
i. x² – 7y + 2 = 0
ii. y² = 5y – 10
iii. y² + \(\frac { 1 }{ y } \) = 2
iv. x + \(\frac { 1 }{ x } \) = -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.

ii. The given equation is
y² = 5y – 10
∴ y² – 5y + 10 = 0
Here, y is the only variable and maximum index of the variable is 2.
a = 1, b = -5, c = 10 are real numbers and a ≠ 0.
∴ The given equation is a quadratic equation.

iii. The given equation is
y² + \(\frac { 1 }{ y } \) = 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.

iv. The given equation is
x + \(\frac { 1 }{ x } \) = -2
∴ x² + 1 = -2x …[Multiplying both sides by x]
∴ x² + 2x+ 1 = 0
Here, x is the only variable and maximum index of the variable is 2.
a = 1, b = 2, c = 1 are real numbers and a ≠ 0.
∴ The given equation is 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.

Question 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

Question 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, \(\frac { 5 }{ 2 } \)
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 = \(\frac { 5 }{ 2 } \) in L.H.S. of equation (i), we get

Question 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 = \(\frac { 27 }{ 9 } \)
∴ k = 3

Question 6.
One of the roots of equation 5m² + 2m + k = 0 is \(\frac { -7 }{ 5 } \) Complete the following activity to find the value of ‘k’.
Solution:

Question 1.
x² + 3x – 5, 3x² – 5x, 5x²; Write the polynomials In the index form. Observe the coefficients and fill in the boxes. (Textbook p. no. 31)
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.

Question 2.
Complete the following table (Textbook p. no. 31)

Question 3.
Decide which of the following are quadratic equations? (Textbook pg. no. 31)
i. 9y² + 5 = 0
ii. m³ – 5m² + 4 = 0
iii. (l + 2)(l – 5) = 0
Solution:
i. 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. 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
∴ 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.

Question 4.
If x = 5 is a root of equation kx² – 14x – 5 = 0, then find the value of k by completing the following activity. (Textbook pg, no. 33)
Solution: