Maharashtra Board Class 10 Maths Chapter 2 Quadratic Equations Set 2.1 Solutions

Question 1. Write any two quadratic equations.
Answer: Solution:
(i) \( y² - 7y + 12 = 0 \)
(ii) \( x² - 8 = 0 \)
In simple words: A quadratic equation is a polynomial equation of the second degree, meaning it contains at least one term in which the unknown variable is squared. Its standard form is \( ax² + bx + c = 0 \), where \( a \), \( b \), and \( c \) are real numbers and \( a \neq 0 \).

🎯 Exam Tip: Remember to always include the equality sign and define the equation to be quadratic. The highest power of the variable must be 2.

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) \( (m + 2) (m - 5) = 0 \)
(vi) \( m³ + 3m² - 2 = 3m³ \)
Answer: 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 \( a \neq 0 \). ∴ 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 \( a \neq 0 \). ∴ The given equation is a quadratic equation.
(iii) The given equation is \( y² + \frac{1}{y} = 2 \)
\( \implies \) \( y³ + 1 = 2y \) ...[Multiplying both sides by y]
\( \implies \) \( y³ - 2y + 1 = 0 \) 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 \( x + \frac{1}{x} = -2 \)
\( \implies \) \( x² + 1 = -2x \) ...[Multiplying both sides by x]
\( \implies \) \( x² + 2x+ 1 = 0 \) Here, x is the only variable and maximum index of the variable is 2. \( a = 1, b = 2, c = 1 \) are real numbers and \( a \neq 0 \). ∴ The given equation is a quadratic equation.
(v) The given equation is \( (m + 2) (m - 5) = 0 \)
\( \implies \) \( m(m - 5) + 2(m - 5) = 0 \)
\( \implies \) \( m² - 5m + 2m - 10 = 0 \)
\( \implies \) \( m² - 3m - 10 = 0 \) Here, m is the only variable and maximum index of the variable is 2. \( a = 1, b = -3, c = -10 \) are real numbers and \( a \neq 0 \). ∴ The given equation is a quadratic equation.
(vi) The given equation is \( m³ + 3m² - 2 = 3m³ \)
\( \implies \) \( 3m³ - m³ - 3m² + 2 = 0 \)
\( \implies \) \( 2m³ - 3m² + 2 = 0 \) Here, m is the only variable and maximum index of the variable is not 2. ∴ The given equation is not a quadratic equation.
In simple words: To decide if an equation is quadratic, check if there's only one variable and its highest power (index) is exactly 2, and the coefficient of the squared term is not zero. If these conditions are met, it's a quadratic equation.

🎯 Exam Tip: Always simplify the equation to its standard form \( ax² + bx + c = 0 \) before determining if it is quadratic. Pay close attention to the highest power of the variable after simplification.

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) \( x² - 9 = 13 \)
Answer: Solution:
(i) \( 2y = 10 - y² \)
\( \implies \) \( y² + 2y - 10 = 0 \) Comparing the above equation with \( ay² + by + c = 0 \), we get \( a = 1, b = 2, c = -10 \)
(ii) \( (x - 1)² = 2x + 3 \)
\( \implies \) \( x² - 2x + 1 = 2x + 3 \)
\( \implies \) \( x² - 2x - 2x + 1 - 3 = 0 \)
\( \implies \) \( x² - 4x - 2 = 0 \) Comparing the above equation with \( ax² + bx + c = 0 \), we get \( a = 1, b = -4, c = -2 \)
(iii) \( x² + 5x = - (3 - x) \)
\( \implies \) \( x² + 5x = -3 + x \)
\( \implies \) \( x² + 5x - x + 3 = 0 \)
\( \implies \) \( x² + 4x + 3 = 0 \) Comparing the above equation with \( ax² + bx + c = 0 \), we get \( a = 1, b = 4, c = 3 \)
(iv) \( 3m² = 2m² - 9 \)
\( \implies \) \( 3m² - 2m² + 9 = 0 \)
\( \implies \) \( m² + 9 = 0 \)
\( \implies \) \( m² + 0m + 9 = 0 \) Comparing the above equation with \( am² + bm + c = 0 \), we get \( a = 1, b = 0, c = 9 \)
(v) \( p (3 + 6p) = - 5 \)
\( \implies \) \( 3p + 6p² = -5 \)
\( \implies \) \( 6p² + 3p + 5 = 0 \) Comparing the above equation with \( ap² + bp + c = 0 \), we get \( a = 6, b = 3, c = 5 \)
(vi) \( x² - 9 = 13 \)
\( \implies \) \( x² - 9 - 13 = 0 \)
\( \implies \) \( x² - 22 = 0 \)
\( \implies \) \( x² + 0x - 22 = 0 \) Comparing the above equation with \( ax² + bx + c = 0 \), we get \( a = 1, b = 0, c = -22 \)
In simple words: To convert any quadratic equation to the standard form \( ax² + bx + c = 0 \), arrange all terms on one side of the equation, setting the other side to zero, and then identify the coefficients \( a \) (of \( x² \)), \( b \) (of \( x \)), and \( c \) (the constant term).

🎯 Exam Tip: Always ensure the coefficient 'a' is positive by multiplying by -1 if needed, though it's not strictly necessary for identifying a, b, c. Make sure to correctly handle signs when moving terms across the equality sign.

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) \( 2m² - 5m = 0 \); \( m = 2, \frac{5}{2} \)
Answer: 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. \( = (-1)² + 4(-1) - 5 = 1 - 4 - 5 = -8 \) ∴ L.H.S. \( \neq \) R.H.S. ∴ \( x = -1 \) is 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. \( \neq \) 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 L.H.S. \( = 2 \left( \frac{5}{2} \right)² - 5 \left( \frac{5}{2} \right) \) \( = 2 \left( \frac{25}{4} \right) - \frac{25}{2} \) \( = \frac{25}{2} - \frac{25}{2} \) \( = 0 \) ∴ L.H.S. = R.H.S. ∴ \( m = \frac{5}{2} \) is the root of the given quadratic equation.
In simple words: To check if a value is a root of an equation, substitute the value into the equation. If both sides of the equation become equal (L.H.S. = R.H.S.), then the value is a root; otherwise, it is not.

🎯 Exam Tip: Be careful with calculations, especially with negative numbers and fractions. A single arithmetic error can lead to an incorrect conclusion about whether a value is a root.

Question 5. Find k if \( x = 3 \) is a root of equation \( kx² - 10x + 3 = 0 \).
Answer: Solution: \( x = 3 \) is the root of the equation \( kx² - 10x + 3 = 0 \). Putting \( x = 3 \) in the given equation, we get \( k(3)² - 10(3) + 3 = 0 \)
\( \implies \) \( 9k - 30 + 3 = 0 \)
\( \implies \) \( 9k - 27 = 0 \)
\( \implies \) \( 9k = 27 \)
\( \implies \) \( k = \frac{27}{9} \)
\( \implies \) \( k = 3 \)
In simple words: If a number is a root of an equation, it means that when you substitute that number in place of the variable, the equation holds true. By substituting the given root, we can solve for the unknown constant 'k'.

🎯 Exam Tip: This type of question tests your understanding of roots and basic algebraic manipulation. Ensure all arithmetic steps, especially multiplication and division, are accurate to find the correct value of 'k'.

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'.
Answer: Solution:
∴ \( \boxed{\frac{-7}{5}} \) is a root of quadratic equation \( 5m² + 2m + k = 0 \).
∴ Put \( m = \boxed{\frac{-7}{5}} \) in the equation.
\( 5 \times \boxed{\left( \frac{-7}{5} \right)²} + 2 \times \boxed{\frac{-7}{5}} + k = 0 \)
∴ \( 5 \times \frac{49}{25} + \left( \frac{-14}{5} \right) + k = 0 \)
∴ \( \boxed{\frac{49}{5}} + \boxed{\frac{-14}{5}} + k = 0 \)
∴ \( \boxed{\frac{35}{5}} + k = 0 \)
∴ \( \boxed{7} + k = 0 \)
∴ \( k = \boxed{-7} \)
In simple words: When a root of an equation is provided, substitute that value into the equation in place of the variable. Then, simplify the equation to solve for any unknown constants, such as 'k' in this problem.

🎯 Exam Tip: For activity-based questions, accurately fill in each blank. Show intermediate steps clearly, especially for calculations involving fractions and squares, to avoid errors and score full marks.

Question 1. \( x² + 3x - 5, 3x2 - 5x, 5x2 \); 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.
In simple words: The index form of a polynomial ensures all terms are present, even if their coefficient is zero, to clearly show the powers of the variable. We then identify the numbers multiplying the variable terms and the constant.

🎯 Exam Tip: When writing polynomials in index form, ensure you include terms with zero coefficients (e.g., \( 0x \), \( 0x² \)) for completeness. Clearly identify \( x² \) coefficients, \( x \) coefficients, and constant terms.

Question 2. Complete the following table

Answer:

In simple words: To complete this table, first convert each given quadratic equation into its standard form \( ax² + bx + c = 0 \). Once in standard form, simply read off the coefficients for \( x² \) (which is \( a \)), \( x \) (which is \( b \)), and the constant term (which is \( c \)).

🎯 Exam Tip: When converting to the general form, ensure all terms are on one side of the equality and the other side is zero. If a term (like \( x \) or constant) is missing, its coefficient is 0.

Question 3. Decide which of the following are quadratic equations?
(i) \( 9y² + 5 = 0 \)
(ii) \( m³ - 5m² + 4 = 0 \)
(iii) \( (l + 2)(1 - 5) = 0 \)
Answer: 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) \( (l + 2)(l - 5) = 0 \)
\( \implies \) \( l(l - 5) + 2(l - 5) = 0 \)
\( \implies \) \( l² - 5l + 2l - 10 = 0 \)
\( \implies \) \( l² - 3l - 10 = 0 \). In this equation [l] is the only variable and maximum index of the variable is [2]. ∴ it [is] a quadratic equation.
In simple words: An equation is quadratic if it has only one variable and the highest power of that variable is 2. If the highest power is 3 or any other number, or if there's more than one variable where one is squared, it's not quadratic.

🎯 Exam Tip: Always simplify and expand equations first. For product forms like \((l+2)(l-5)\), multiply them out to reveal the highest power of the variable before deciding if it's quadratic.

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.
Answer: Solution:
One of the roots of equation \( kx² - 14x - 5 = 0 \) is \( \boxed{5} \)
∴ Put \( x = \boxed{5} \) in the given equation.
\( k \boxed{5²} - 14 \boxed{(5)} - 5 = 0 \)
∴ \( 25k - 70 - 5 = 0 \)
∴ \( 25k - \boxed{75} = 0 \)
∴ \( 25k = \boxed{75} \)
∴ \( k = \frac{\boxed{75}}{\boxed{25}} \)
∴ \( k = \boxed{3} \)
In simple words: When a value is a root of an equation, it satisfies the equation. By substituting the given root into the equation, we can form a solvable linear equation for the unknown constant 'k'.

🎯 Exam Tip: Carefully substitute the value of the root into the equation. Pay close attention to signs and order of operations (e.g., squaring before multiplying) to avoid calculation errors in activity-based questions.