Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3.11

Question 1. Solve the following quadratic equations by completing the square method
(i) \( 9x^2 - 12x + 4 = 0 \)
Answer:
We are given the quadratic equation: \( 9x^2 - 12x + 4 = 0 \)
First, we divide the entire equation by 9 to make the coefficient of \( x^2 \) equal to 1:
\( \frac{9x^2}{9} - \frac{12x}{9} + \frac{4}{9} = 0 \)
\( x^2 - \frac{4x}{3} + \frac{4}{9} = 0 \)
Now, we move the constant term to the right side:
\( x^2 - \frac{4x}{3} = - \frac{4}{9} \)
To complete the square, we need to add \( (\frac{1}{2} \times \text{coefficient of } x)^2 \) to both sides.
The coefficient of \( x \) is \( -\frac{4}{3} \).
So, \( (\frac{1}{2} \times -\frac{4}{3})^2 = (-\frac{4}{6})^2 = (-\frac{2}{3})^2 \)
Add \( (-\frac{2}{3})^2 \) to both sides:
\( x^2 - \frac{4x}{3} + (-\frac{2}{3})^2 = -\frac{4}{9} + (-\frac{2}{3})^2 \)
\( (x - \frac{2}{3})^2 = -\frac{4}{9} + \frac{4}{9} \)
\( (x - \frac{2}{3})^2 = 0 \)
Take the square root of both sides:
\( x - \frac{2}{3} = 0 \)
\( x = \frac{2}{3} \)
This means the quadratic equation has a single, repeated real root. This happens when the discriminant is zero.
The solution is \( \frac{2}{3} \).
In simple words: To solve this, we first change the equation so \( x^2 \) stands alone. Then, we add a special number to both sides to make the left side a perfect square, like \( (a-b)^2 \). After that, we find the value of \( x \) by taking the square root.

🎯 Exam Tip: When completing the square, remember to always divide by the coefficient of \( x^2 \) first. Also, ensure you add \( (\text{half the coefficient of } x)^2 \) to both sides of the equation.

Question 1. (ii) \( \frac { 5x+7 }{x-1} = 3x + 2 \)
Answer:
We are given the equation: \( \frac{5x+7}{x-1} = 3x + 2 \)
First, multiply both sides by \( (x-1) \) to remove the fraction:
\( 5x + 7 = (3x + 2)(x - 1) \)
Now, expand the right side:
\( 5x + 7 = 3x(x - 1) + 2(x - 1) \)
\( 5x + 7 = 3x^2 - 3x + 2x - 2 \)
Combine like terms on the right side:
\( 5x + 7 = 3x^2 - x - 2 \)
Move all terms to one side to form a standard quadratic equation \( ax^2 + bx + c = 0 \):
\( 0 = 3x^2 - x - 5x - 2 - 7 \)
\( 3x^2 - 6x - 9 = 0 \)
Divide the entire equation by 3 to simplify:
\( \frac{3x^2}{3} - \frac{6x}{3} - \frac{9}{3} = 0 \)
\( x^2 - 2x - 3 = 0 \)
Now, we will complete the square. Move the constant term to the right side:
\( x^2 - 2x = 3 \)
To complete the square, add \( (\frac{1}{2} \times \text{coefficient of } x)^2 \) to both sides.
The coefficient of \( x \) is \( -2 \).
So, \( (\frac{1}{2} \times -2)^2 = (-1)^2 = 1 \)
Add 1 to both sides:
\( x^2 - 2x + 1 = 3 + 1 \)
The left side is now a perfect square:
\( (x - 1)^2 = 4 \)
Take the square root of both sides:
\( x - 1 = \pm \sqrt{4} \)
\( x - 1 = \pm 2 \)
This gives us two possible cases:
Case 1: \( x - 1 = 2 \)
\( x = 2 + 1 \)
\( x = 3 \)
Case 2: \( x - 1 = -2 \)
\( x = -2 + 1 \)
\( x = -1 \)
The solution set is \( \{-1, 3\} \). This method helps find roots by changing the equation into a perfect square form.
In simple words: First, we get rid of the fraction by multiplying. Then, we arrange the equation into a normal quadratic form. After that, we use the "completing the square" trick to find the values of \( x \) that make the equation true.

🎯 Exam Tip: When dealing with fractional equations, always clear the denominators first. Remember that taking the square root of a number like 4 gives both \( +2 \) and \( -2 \), leading to two possible solutions.

Question 2. Solve the following quadratic equations by formula method
(i) (Equation implied from coefficients `a = 2, b = -5, c = 2`)
Answer:
The given quadratic equation has coefficients: \( a = 2, b = -5, c = 2 \)
We use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Substitute the values of \( a, b, \) and \( c \) into the formula:
\( x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)} \)
\( x = \frac{5 \pm \sqrt{25 - 16}}{4} \)
\( x = \frac{5 \pm \sqrt{9}}{4} \)
\( x = \frac{5 \pm 3}{4} \)
Now, we find the two possible solutions:
First solution: \( x = \frac{5 + 3}{4} \)
\( x = \frac{8}{4} \)
\( x = 2 \)
Second solution: \( x = \frac{5 - 3}{4} \)
\( x = \frac{2}{4} \)
\( x = \frac{1}{2} \)
The solution set is \( \{\frac{1}{2}, 2\} \). The quadratic formula is very useful for solving any quadratic equation, even if it cannot be easily factored.
In simple words: We use a special formula that directly gives the answers for \( x \). We just put in the numbers \( a, b, \) and \( c \) from the equation into the formula and do the math to get the two solutions.

🎯 Exam Tip: Always write down the values of \( a, b, \) and \( c \) clearly before substituting them into the quadratic formula to avoid sign errors, especially with negative values.

Question 2. (ii) \( \sqrt { 2 } f^2 – 6 f + 3 \sqrt { 2 } = 0 \)
Answer:
The given quadratic equation is: \( \sqrt{2} f^2 - 6f + 3\sqrt{2} = 0 \)
Comparing this with \( af^2 + bf + c = 0 \), we have:
\( a = \sqrt{2} \)
\( b = -6 \)
\( c = 3\sqrt{2} \)
Using the quadratic formula: \( f = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Substitute the values of \( a, b, \) and \( c \):
\( f = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(\sqrt{2})(3\sqrt{2})}}{2(\sqrt{2})} \)
\( f = \frac{6 \pm \sqrt{36 - 4(3 \times 2)}}{2\sqrt{2}} \)
\( f = \frac{6 \pm \sqrt{36 - 4(6)}}{2\sqrt{2}} \)
\( f = \frac{6 \pm \sqrt{36 - 24}}{2\sqrt{2}} \)
\( f = \frac{6 \pm \sqrt{12}}{2\sqrt{2}} \)
We can simplify \( \sqrt{12} \) as \( \sqrt{4 \times 3} = 2\sqrt{3} \):
\( f = \frac{6 \pm 2\sqrt{3}}{2\sqrt{2}} \)
Factor out 2 from the numerator:
\( f = \frac{2(3 \pm \sqrt{3})}{2\sqrt{2}} \)
Cancel out the 2:
\( f = \frac{3 \pm \sqrt{3}}{\sqrt{2}} \)
The two solutions are:
\( f_1 = \frac{3 + \sqrt{3}}{\sqrt{2}} \)
\( f_2 = \frac{3 - \sqrt{3}}{\sqrt{2}} \)
These solutions represent the points where the quadratic function crosses the f-axis.
In simple words: For this equation, we find \( a, b, \) and \( c \) which have square roots. Then, we put these into the quadratic formula. We simplify the square root part and then divide to get the two different answers for \( f \).

🎯 Exam Tip: Be careful when multiplying square roots (e.g., \( \sqrt{2} \times \sqrt{2} = 2 \)) and simplifying terms within the discriminant, \( b^2 - 4ac \). Always simplify the square root of the discriminant as much as possible before the final step.

Question 2. (iii) \( 3y^2 – 20y – 23 = 0 \)
Answer:
The given quadratic equation is: \( 3y^2 - 20y - 23 = 0 \)
Comparing this with \( ay^2 + by + c = 0 \), we have:
\( a = 3 \)
\( b = -20 \)
\( c = -23 \)
Using the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Substitute the values of \( a, b, \) and \( c \):
\( y = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(3)(-23)}}{2(3)} \)
\( y = \frac{20 \pm \sqrt{400 + 276}}{6} \)
\( y = \frac{20 \pm \sqrt{676}}{6} \)
We know that \( \sqrt{676} = 26 \).
\( y = \frac{20 \pm 26}{6} \)
Now, we find the two possible solutions:
First solution: \( y = \frac{20 + 26}{6} \)
\( y = \frac{46}{6} \)
\( y = \frac{23}{3} \)
Second solution: \( y = \frac{20 - 26}{6} \)
\( y = \frac{-6}{6} \)
\( y = -1 \)
The solution set is \( \{-1, \frac{23}{3}\} \). The quadratic formula is a direct way to find these solutions without trial and error.
In simple words: We take the numbers \( a, b, \) and \( c \) from the equation and put them into the quadratic formula. After doing the calculations, we find two answers for \( y \).

🎯 Exam Tip: Pay close attention to negative signs, especially when \( c \) is negative, as \( -4ac \) will become positive. Memorizing common squares and their square roots (like \( \sqrt{676} = 26 \)) can save time.

Question 2. (iv) \( 36y^2 – 12ay + (a^2 – b^2) = 0 \)
Answer:
The given quadratic equation is: \( 36y^2 - 12ay + (a^2 - b^2) = 0 \)
Comparing this with \( Ay^2 + By + C = 0 \), we have:
\( A = 36 \)
\( B = -12a \)
\( C = a^2 - b^2 \)
Using the quadratic formula: \( y = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \)
Substitute the values of \( A, B, \) and \( C \):
\( y = \frac{-(-12a) \pm \sqrt{(-12a)^2 - 4(36)(a^2 - b^2)}}{2(36)} \)
\( y = \frac{12a \pm \sqrt{144a^2 - 144(a^2 - b^2)}}{72} \)
\( y = \frac{12a \pm \sqrt{144a^2 - 144a^2 + 144b^2}}{72} \)
\( y = \frac{12a \pm \sqrt{144b^2}}{72} \)
\( y = \frac{12a \pm 12b}{72} \)
Now, we find the two possible solutions:
First solution: \( y = \frac{12a + 12b}{72} \)
Factor out 12 from the numerator:
\( y = \frac{12(a + b)}{72} \)
Cancel out 12:
\( y = \frac{a + b}{6} \)
Second solution: \( y = \frac{12a - 12b}{72} \)
Factor out 12 from the numerator:
\( y = \frac{12(a - b)}{72} \)
Cancel out 12:
\( y = \frac{a - b}{6} \)
The solution set is \( \{\frac{a+b}{6}, \frac{a-b}{6}\} \). This shows how the formula can solve equations with variables in the coefficients.
In simple words: Here, the numbers \( a, b, \) and \( c \) themselves contain other letters. We still use the same quadratic formula, but we have to be careful with the algebra when we put in these values. In the end, we get two answers for \( y \) that are also expressions with letters.

🎯 Exam Tip: When coefficients involve variables, treat them as constants in the formula. Remember to expand and simplify terms carefully, especially when squaring algebraic expressions like \( (-12a)^2 \).

Question 3. Find the time when the distance travelled by the ball is 11.25 feet.
Answer:
The distance \( d \) travelled by the ball is given by the formula: \( d = t^2 - 0.75t \)
We are given that the distance travelled is 11.25 feet. So, we set \( d = 11.25 \):
\( 11.25 = t^2 - 0.75t \)
To solve this quadratic equation, first, we move all terms to one side:
\( t^2 - 0.75t - 11.25 = 0 \)
To work with whole numbers, we multiply the entire equation by 100:
\( 100(t^2 - 0.75t - 11.25) = 100(0) \)
\( 100t^2 - 75t - 1125 = 0 \)
We can simplify this equation by dividing by 25:
\( \frac{100t^2}{25} - \frac{75t}{25} - \frac{1125}{25} = \frac{0}{25} \)
\( 4t^2 - 3t - 45 = 0 \)
Now, we use the quadratic formula to solve for \( t \). The coefficients are:
\( a = 4 \)
\( b = -3 \)
\( c = -45 \)
The quadratic formula is: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Substitute the values of \( a, b, \) and \( c \):
\( t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(4)(-45)}}{2(4)} \)
\( t = \frac{3 \pm \sqrt{9 - (-720)}}{8} \)
\( t = \frac{3 \pm \sqrt{9 + 720}}{8} \)
\( t = \frac{3 \pm \sqrt{729}}{8} \)
We know that \( \sqrt{729} = 27 \).
\( t = \frac{3 \pm 27}{8} \)
Now, we find the two possible solutions for \( t \):
First solution: \( t = \frac{3 + 27}{8} \)
\( t = \frac{30}{8} \)
\( t = \frac{15}{4} \)
\( t = 3.75 \) seconds
Second solution: \( t = \frac{3 - 27}{8} \)
\( t = \frac{-24}{8} \)
\( t = -3 \) seconds
Since time cannot be negative, we discard the second solution \( t = -3 \). The final answer must make sense in the real world.
Therefore, the required time is \( \frac{15}{4} \) seconds or 3.75 seconds.
In simple words: We are given a formula for distance and a total distance. We put these into an equation and solve it for time. Since time cannot be a negative number, we choose the positive answer.

🎯 Exam Tip: For word problems involving physical quantities like time, distance, or length, always check if your solutions make sense in the real world. Negative values for such quantities are usually discarded unless the context specifies otherwise.