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 โ€“ 7y + 12 = 0
ii. x2 โ€“ 8 = 0

Question 2.
Decide which of the following are quadratic
i. x2 โ€“ 7y + 2 = 0
ii. y2 = 5y โ€“ 10
iii. y2 + \(\frac { 1 }{ y } \) = 2
iv. x + \(\frac { 1 }{ x } \) = -2
v. (m + 2) (m โ€“ 5) = 03
vi. m3 + 3m2 โ€“ 2 = 3m3
Solution:
i. The given equation is x2 + 5x โ€“ 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 โ‰  0.
โˆด The given equation is a quadratic equation.

ii. The given equation is
y2 = 5y โ€“ 10
โˆด y2 โ€“ 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 โ‰  0.
โˆด The given equation is a quadratic equation.

iii. The given equation is
y2 + \(\frac { 1 }{ y } \) = 2
โˆด y3 + 1 = 2y โ€ฆ[Multiplying both sides by y]
โˆด y3 โ€“ 2y + 1 = 0
Here, y is the only variable and maximum index of the variable is not 2.
โˆด The given equation is not a quadratic equation.

iv. The given equation is
x + \(\frac { 1 }{ x } \) = -2
โˆด x2 + 1 = -2x โ€ฆ[Multiplying both sides by x]
โˆด 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 โ‰  0.
โˆด The given equation is a quadratic equation.

v. The given equation is
(m + 2) (m โ€“ 5) = 0
โˆด m(m โ€“ 5) + 2(m โ€“ 5) = 0
โˆด m2 โ€“ 5m + 2m โ€“ 10 = 0
โˆด m2 โ€“ 3m โ€“ 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 โ‰  0.
โˆด The given equation is a quadratic equation.

vi. The given equation is
m3 + 3m2 โ€“ 2 = 3m3
โˆด 3m3 โ€“ m3 โ€“ 3m2 + 2 = 0
โˆด 2m3 โ€“ 3m2 + 2 = 0
Here, m is the only variable and maximum
index of the variable is not 2.
โˆด 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 โ€“ y2
ii. (x โ€“ 1)2 = 2x + 3
iii. x2 + 5x = โ€“ (3 โ€“ x)
iv. 3m2 = 2m2 โ€“ 9
v. P (3 + 6p) = โ€“ 5
vi. x2 โ€“ 9 = 13
Solution:
i. 2y โ€“ 10 โ€“ y2
โˆด y2 + 2y โ€“ 10 = 0
Comparing the above equation with
ay2 + by + c = 0, we get
a = 1, b = 2, c = -10

ii. (x โ€“ 1)2 = 2x + 3
โˆด x2 โ€“ 2x + 12x + 3
x2 โ€“ 2x + 1 โ€“ 2x โ€“ 30
โˆด x2 โ€“ 4x โ€“ 2 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = 1, b = -4, c = -2

iii. x2 + 5x = โ€“ (3 โ€“ x)
โˆด x2 + 5x = -3 + x
โˆด x2 + 5x โ€“ x + 3 = 0
โˆด x2 + 4x + 3 = 0
Comparing the above equation with
ax2 + bx + c = 0, we get
a = 1, b = 4, c = 3

iv. 3m2 = 2m2 โ€“ 9
โˆด 3m2 โ€“ 2m2 + 9 = 0
โˆด m2 + 9 = 0
โˆด 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) = โ€“ 5
โˆด 3p + 6p2 = -5
โˆด 6p2 + 3p + 5 = 0
Comparing the above equation with
ap2 + bp + c = 0, we get
a = 6, b = 3, c = 5

vi. x2 โ€“ 9 = 13
โˆด x2 โ€“ 9 โ€“ 13 = 0
โˆด x2 โ€“ 22 = 0
โˆด x2 + 0x โ€“ 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 โ€“ 5 = 0; x = 1,-1
ii. 2m2 โ€“ 5m = 0; m = 2, \(\frac { 5 }{ 2 } \)
Solution:
i. The given equation is
x2 + 4x โ€“ 5 = 0 โ€ฆ(i)
Putting x = 1 in L.H.S. of equation (i), we get
L.H.S. = (1)2 + 4(1) โ€“ 5 = 1 + 4 โ€“ 5 = 0
โˆด L.H.S. = R.H.S.
โˆด 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) โ€“ 5 = 1 โ€“ 4 โ€“ 5 = -8
โˆด LH.S. โ‰  R.H.S.
โˆด x = -1 ยกs not the root of the given quadratic equation.

ii. The given equation is
2m2 โ€“ 5m = 0 โ€ฆ(i)
Putting m = 2 in L.H.S. of equation (i), we get
L.H.S. = 2(2)2 โ€“ 5(2) = 2(4) -10 = 8 โ€“ 10 = -2
โˆด L.H.S. โ‰  R.H.S.
โˆด 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 โ€“ 10x + 3 = 0.
Solution:
x = 3 is the root of the equation kx2 โ€“ 10x + 3 = 0.
Putting x = 3 in the given equation, we get
k(3)2 โ€“ 10(3) + 3 = 0
โˆด 9k โ€“ 30 +3 = 0
โˆด 9k โ€“ 27 = 0
โˆด 9k = 27
โˆด k = \(\frac { 27 }{ 9 } \)
โˆด 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 โ€˜kโ€™.
Solution:


Question 1.
x2 + 3x โ€“ 5, 3x2 โ€“ 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 โ€“ 5, 3x2 โ€“ 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)

Answer:


Question 3.
Decide which of the following are quadratic equations? (Textbook pg. no. 31)
i. 9y2 + 5 = 0
ii. m3 โ€“ 5m2 + 4 = 0
iii. (l + 2)(l โ€“ 5) = 0
Solution:
i. In the equation 9y2 + 5 = 0, [y] is the only variable and maximum index of the variable is [2].
โˆด It [is] a quadratic equation.

ii. In the equation m3 โ€“ 5m2 + 4 = 0, [m] is the only variable and maximum index of the variable is not 2.
โˆด It [is not] a quadratic equation.

iii. (l + 2)(l โ€“ 5) = 0
โˆด l(l โ€“ 5) + 2(l โ€“ 5) = 0
โˆด l2 โ€“ 5l + 2l โ€“ 10 = 0
โˆด l2 โ€“ 3l โ€“ 10 = 0.
In this equation [l] is the only variable and maximum index of the variable is [2]
โˆด it [is] a quadratic equation.

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