OP Malhotra Class 10 Maths Solutions Chapter 5 Quadratic Equations Exercise 5 (A)

Question 1. (i) \( (x - 3)(x + 7) = 0 \)
Answer: We need to find the values of \( x \) that make this equation true. Since two terms multiplied together result in zero, one of them must be zero. If the first term, \( (x-3) \), is equal to zero, then \( x \) must be 3. Alternatively, if the second term, \( (x+7) \), is equal to zero, then \( x \) must be -7. Therefore, the solutions for \( x \) are 3 and -7.
In simple words: To solve this, make each part of the multiplication equal to zero. If \( x-3=0 \), then \( x=3 \). If \( x+7=0 \), then \( x=-7 \). These are the two answers for \( x \).

🎯 Exam Tip: When an equation is already factored and set to zero, simply set each factor equal to zero to find the solutions.

Question 1. (ii) \( (3x + 4) (2x - 11) = 0 \)
Answer: Similar to the previous part, we set each factor in the equation equal to zero to find the values of \( x \). If \( 3x + 4 = 0 \), then \( 3x = -4 \), which means \( x = -\frac{4}{3} \). If \( 2x - 11 = 0 \), then \( 2x = 11 \), leading to \( x = \frac{11}{2} \). These are the two possible values for \( x \).
In simple words: For this equation, make each bracket equal to zero. From \( 3x+4=0 \), we get \( x = -4/3 \). From \( 2x-11=0 \), we get \( x = 11/2 \). These are the answers.

🎯 Exam Tip: Remember to solve for \( x \) completely after setting each factor to zero, including any division by coefficients.

Question 2. \( x^2 = 4x \)
Answer: To solve this quadratic equation, first rearrange it so all terms are on one side, making it equal to zero: \( x^2 - 4x = 0 \). Next, factor out the common term, \( x \), to get \( x(x-4) = 0 \). For this product to be zero, either \( x \) itself must be zero, or the term \( (x-4) \) must be zero, which gives \( x=4 \). The solutions are \( x = 0 \) or \( x = 4 \). Quadratic equations can often have two distinct solutions.
In simple words: First, move \( 4x \) to the other side to get \( x^2 - 4x = 0 \). Then, take out the common \( x \), so it becomes \( x(x-4) = 0 \). This means either \( x \) is 0, or \( x-4 \) is 0 (which means \( x=4 \)). So, \( x \) can be 0 or 4.

🎯 Exam Tip: Never divide by a variable (like \( x \)) if it might be zero, as you could lose a valid solution. Always move all terms to one side and factor.

Question 3. \( \left(\frac{1}{3} x-1\right)\left(\frac{1}{2} x+7\right) = 0 \)
Answer: For the product of two factors to be zero, at least one of the factors must be zero. Setting the first factor, \( \left(\frac{1}{3}x - 1\right) \), to zero, we get \( \frac{1}{3}x = 1 \), which implies \( x = 3 \). Setting the second factor, \( \left(\frac{1}{2}x + 7\right) \), to zero, we find \( \frac{1}{2}x = -7 \), which simplifies to \( x = -14 \). Thus, the solutions are \( x = 3 \) and \( x = -14 \). This method is widely used for solving factored polynomial equations.
In simple words: We have two parts multiplied together that equal zero. So, we set each part to zero. If \( \frac{1}{3}x - 1 = 0 \), then \( x = 3 \). If \( \frac{1}{2}x + 7 = 0 \), then \( x = -14 \). These are the two answers for \( x \).

🎯 Exam Tip: Handle fractions carefully when solving for \( x \) after setting each factor to zero; multiply by the reciprocal to isolate \( x \).

Question 4. \( \frac{x^2-5 x}{2} = 0 \)
Answer: To solve this equation, we first multiply both sides by 2 to remove the denominator, resulting in \( x^2 - 5x = 0 \). Then, we factor out the common term \( x \), giving us \( x(x-5) = 0 \). This means either \( x \) is 0 or \( x-5 \) is 0, which makes \( x=5 \). The two solutions for \( x \) are 0 and 5. Clearing the denominator is a key first step in solving fractional equations.
In simple words: First, get rid of the `2` under the fraction by multiplying both sides by 2, leaving \( x^2 - 5x = 0 \). Then, take out the common \( x \), so \( x(x-5)=0 \). This means \( x \) is 0, or \( x-5 \) is 0, so \( x=5 \). The answers are 0 and 5.

🎯 Exam Tip: When a fraction equals zero, only the numerator needs to be zero, assuming the denominator is not zero.

Question 5. \( x^2 - 3x - 10 = 0 \)
Answer: This quadratic equation can be solved by factoring. We need two numbers that multiply to -10 and add up to -3. These numbers are -5 and 2. So, we rewrite the middle term: \( x^2 - 5x + 2x - 10 = 0 \). Grouping terms and factoring, we get \( x(x-5) + 2(x-5) = 0 \), which simplifies to \( (x-5)(x+2) = 0 \). Setting each factor to zero gives \( x=5 \) or \( x=-2 \). Factoring by splitting the middle term is a common method for non-monic quadratics.
In simple words: To solve this, we find two numbers that multiply to -10 and add to -3 (these are -5 and 2). We then split the middle term: \( x^2 - 5x + 2x - 10 = 0 \). Group the terms: \( x(x-5) + 2(x-5) = 0 \). This simplifies to \( (x-5)(x+2) = 0 \). So, \( x \) is 5 or \( x \) is -2.

🎯 Exam Tip: Always double-check your factors by multiplying them out to ensure they return the original quadratic equation.

Question 6. \( x^2 + x - 12 = 0 \)
Answer: We solve this quadratic equation by factoring. We look for two numbers that multiply to -12 and add up to 1. These numbers are 4 and -3. We rewrite the middle term \( x \) as \( 4x - 3x \). Then, \( x^2 + 4x - 3x - 12 = 0 \). Factor by grouping: \( x(x+4) - 3(x+4) = 0 \), which leads to \( (x+4)(x-3) = 0 \). Setting each factor to zero gives the solutions \( x=-4 \) or \( x=3 \). Factoring provides a clear path to solutions for many quadratic equations.
In simple words: Find two numbers that multiply to -12 and add to 1 (which are 4 and -3). Split the middle term: \( x^2 + 4x - 3x - 12 = 0 \). Group and factor: \( x(x+4) - 3(x+4) = 0 \), so \( (x+4)(x-3) = 0 \). This means \( x \) is -4 or \( x \) is 3.

🎯 Exam Tip: When splitting the middle term, ensure the sum of the two new coefficients equals the original middle coefficient, and their product equals the product of the first and last coefficients.

Question 7. \( 2 (x^2 + 1) = 5x \)
Answer: First, expand the equation and rearrange it into standard quadratic form: \( 2x^2 + 2 = 5x \) becomes \( 2x^2 - 5x + 2 = 0 \). To factor this, we need two numbers that multiply to \( (2 \times 2 = 4) \) and add to -5. These numbers are -4 and -1. Rewrite the middle term: \( 2x^2 - 4x - x + 2 = 0 \). Factor by grouping: \( 2x(x-2) - 1(x-2) = 0 \), which gives \( (x-2)(2x-1) = 0 \). Setting each factor to zero, we find \( x=2 \) or \( x=\frac{1}{2} \). Always simplify fractions in your final answer.
In simple words: Expand and arrange: \( 2x^2 - 5x + 2 = 0 \). Find two numbers that multiply to 4 (from 2x2) and add to -5 (which are -4 and -1). Split the middle term: \( 2x^2 - 4x - x + 2 = 0 \). Group: \( 2x(x-2) - 1(x-2) = 0 \). So, \( (x-2)(2x-1) = 0 \). This gives \( x=2 \) or \( x=1/2 \).

🎯 Exam Tip: Remember the order of operations: distribute first, then move all terms to one side to get the standard \( ax^2+bx+c=0 \) form.

Question 8. \( x (2x + 5) = 3 \)
Answer: First, expand and rearrange the equation to get it in standard quadratic form: \( 2x^2 + 5x = 3 \) becomes \( 2x^2 + 5x - 3 = 0 \). To factor this, we need two numbers that multiply to \( (2 \times -3 = -6) \) and add to 5. These numbers are 6 and -1. We rewrite the middle term \( 5x \) as \( 6x - x \). Then, \( 2x^2 + 6x - x - 3 = 0 \). Factor by grouping: \( 2x(x+3) - 1(x+3) = 0 \), which results in \( (x+3)(2x-1) = 0 \). Setting each factor to zero, we find the solutions \( x=-3 \) or \( x=\frac{1}{2} \). Proper sign handling during factoring is critical.
In simple words: Expand and arrange: \( 2x^2 + 5x - 3 = 0 \). Find two numbers that multiply to -6 (from 2x-3) and add to 5 (which are 6 and -1). Split the middle term: \( 2x^2 + 6x - x - 3 = 0 \). Group: \( 2x(x+3) - 1(x+3) = 0 \). So, \( (x+3)(2x-1) = 0 \). This gives \( x=-3 \) or \( x=1/2 \).

🎯 Exam Tip: Always make sure the quadratic equation is set to zero before attempting to factor it, otherwise the zero product property cannot be applied.

Question 9. \( 4x^2 - 3x - 1 = 0 \)
Answer: We solve this quadratic equation by factoring. We need two numbers that multiply to \( (4 \times -1 = -4) \) and add up to -3. These numbers are -4 and 1. We rewrite the middle term \( -3x \) as \( -4x + x \). Then, \( 4x^2 - 4x + x - 1 = 0 \). Factor by grouping: \( 4x(x-1) + 1(x-1) = 0 \), which gives \( (x-1)(4x+1) = 0 \). Setting each factor to zero, we find the solutions \( x=1 \) or \( x=-\frac{1}{4} \). Factoring by grouping works well when the leading coefficient is not 1.
In simple words: Find two numbers that multiply to -4 (from 4x-1) and add to -3 (which are -4 and 1). Split the middle term: \( 4x^2 - 4x + x - 1 = 0 \). Group: \( 4x(x-1) + 1(x-1) = 0 \). So, \( (x-1)(4x+1) = 0 \). This means \( x=1 \) or \( x=-1/4 \).

🎯 Exam Tip: When factoring by grouping, ensure that the expressions in the parentheses are identical after factoring out the common terms.

Question 10. \( 6x^2 - 13x + 5 = 0 \)
Answer: We solve this quadratic equation by factoring. We need two numbers that multiply to \( (6 \times 5 = 30) \) and add up to -13. These numbers are -3 and -10. We rewrite the middle term \( -13x \) as \( -3x - 10x \). Then, \( 6x^2 - 3x - 10x + 5 = 0 \). Factor by grouping: \( 3x(2x-1) - 5(2x-1) = 0 \), which results in \( (2x-1)(3x-5) = 0 \). Setting each factor to zero, we find the solutions \( x=\frac{1}{2} \) or \( x=\frac{5}{3} \). Finding the correct pair of numbers to split the middle term is the crucial first step.
In simple words: Find two numbers that multiply to 30 (from 6x5) and add to -13 (which are -3 and -10). Split the middle term: \( 6x^2 - 3x - 10x + 5 = 0 \). Group: \( 3x(2x-1) - 5(2x-1) = 0 \). So, \( (2x-1)(3x-5) = 0 \). This gives \( x=1/2 \) or \( x=5/3 \).

🎯 Exam Tip: Practice identifying factors quickly to speed up the factoring process. This improves with understanding prime factorization.

Question 11. \( 3x^2 - 5x - 12 = 0 \)
Answer: To solve this quadratic equation by factoring, we look for two numbers that multiply to \( (3 \times -12 = -36) \) and add up to -5. These numbers are -9 and 4. We rewrite the middle term \( -5x \) as \( -9x + 4x \). Then, \( 3x^2 - 9x + 4x - 12 = 0 \). Factor by grouping: \( 3x(x-3) + 4(x-3) = 0 \), which results in \( (x-3)(3x+4) = 0 \). Setting each factor to zero, we find the solutions \( x=3 \) or \( x=-\frac{4}{3} \). Factoring is an efficient way to solve quadratics when applicable.
In simple words: Find two numbers that multiply to -36 (from 3x-12) and add to -5 (which are -9 and 4). Split the middle term: \( 3x^2 - 9x + 4x - 12 = 0 \). Group: \( 3x(x-3) + 4(x-3) = 0 \). So, \( (x-3)(3x+4) = 0 \). This gives \( x=3 \) or \( x=-4/3 \).

🎯 Exam Tip: Always make sure to combine like terms correctly before factoring, especially if the equation isn't initially in standard form.

Question 12. \( 2x^2 - 11x + 5 = 0 \)
Answer: To solve this quadratic equation by factoring, we look for two numbers that multiply to \( (2 \times 5 = 10) \) and add up to -11. These numbers are -10 and -1. We rewrite the middle term \( -11x \) as \( -10x - x \). Then, \( 2x^2 - 10x - x + 5 = 0 \). Factor by grouping: \( 2x(x-5) - 1(x-5) = 0 \), which results in \( (x-5)(2x-1) = 0 \). Setting each factor to zero, we find the solutions \( x=5 \) or \( x=\frac{1}{2} \). The method of splitting the middle term relies on finding the correct pair of integers.
In simple words: Find two numbers that multiply to 10 (from 2x5) and add to -11 (which are -10 and -1). Split the middle term: \( 2x^2 - 10x - x + 5 = 0 \). Group: \( 2x(x-5) - 1(x-5) = 0 \). So, \( (x-5)(2x-1) = 0 \). This gives \( x=5 \) or \( x=1/2 \).

🎯 Exam Tip: If the leading coefficient is greater than 1, remember to multiply it by the constant term when looking for factors to split the middle term.

Question 13. \( \frac{x}{2} + \frac{6}{x} = 4 \)
Answer: First, multiply the entire equation by the common denominator \( 2x \) to eliminate fractions. This gives \( x^2 + 12 = 8x \). Rearrange into standard quadratic form: \( x^2 - 8x + 12 = 0 \). To factor, find two numbers that multiply to 12 and add to -8. These numbers are -2 and -6. Rewrite the middle term: \( x^2 - 2x - 6x + 12 = 0 \). Factor by grouping: \( x(x-2) - 6(x-2) = 0 \), which simplifies to \( (x-2)(x-6) = 0 \). Setting each factor to zero, we get \( x=2 \) or \( x=6 \). Always check that the denominators are not zero for your solutions.
In simple words: Multiply the whole equation by \( 2x \) to get rid of fractions: \( x^2 + 12 = 8x \). Move `8x` to the left: \( x^2 - 8x + 12 = 0 \). Find two numbers that multiply to 12 and add to -8 (which are -2 and -6). Split the middle term: \( x^2 - 2x - 6x + 12 = 0 \). Group: \( x(x-2) - 6(x-2) = 0 \). So, \( (x-2)(x-6) = 0 \). This means \( x=2 \) or \( x=6 \).

🎯 Exam Tip: When working with rational equations, it's crucial to identify any values of \( x \) that would make the original denominators zero, as these are invalid solutions.

Question 14. \( 10x - \frac{1}{x} = 3 \)
Answer: First, multiply the entire equation by \( x \) to eliminate the fraction, giving \( 10x^2 - 1 = 3x \). Rearrange into standard quadratic form: \( 10x^2 - 3x - 1 = 0 \). To factor, find two numbers that multiply to \( (10 \times -1 = -10) \) and add to -3. These numbers are -5 and 2. Rewrite the middle term: \( 10x^2 - 5x + 2x - 1 = 0 \). Factor by grouping: \( 5x(2x-1) + 1(2x-1) = 0 \), which simplifies to \( (2x-1)(5x+1) = 0 \). Setting each factor to zero, we get \( x=\frac{1}{2} \) or \( x=-\frac{1}{5} \). Multiplying by the variable \( x \) is a common step for these types of equations.
In simple words: Multiply the whole equation by \( x \) to remove the fraction: \( 10x^2 - 1 = 3x \). Move `3x` to the left: \( 10x^2 - 3x - 1 = 0 \). Find two numbers that multiply to -10 (from 10x-1) and add to -3 (which are -5 and 2). Split the middle term: \( 10x^2 - 5x + 2x - 1 = 0 \). Group: \( 5x(2x-1) + 1(2x-1) = 0 \). So, \( (2x-1)(5x+1) = 0 \). This gives \( x=1/2 \) or \( x=-1/5 \).

🎯 Exam Tip: Be mindful that \( x \) cannot be zero in the original equation. Ensure your final solutions do not violate this condition.

Question 15. \( 9x + \frac{1}{x} = 6 \)
Answer: First, multiply the entire equation by \( x \) to eliminate the fraction: \( 9x^2 + 1 = 6x \). Rearrange into standard quadratic form: \( 9x^2 - 6x + 1 = 0 \). This is a perfect square trinomial, which can be factored as \( (3x-1)^2 = 0 \). Alternatively, to factor, find two numbers that multiply to \( (9 \times 1 = 9) \) and add to -6. These numbers are -3 and -3. Rewrite the middle term: \( 9x^2 - 3x - 3x + 1 = 0 \). Factor by grouping: \( 3x(3x-1) - 1(3x-1) = 0 \), which simplifies to \( (3x-1)(3x-1) = 0 \). Setting the factor to zero, we find \( x=\frac{1}{3} \). This equation has a repeated root, which is common for perfect square quadratics.
In simple words: Multiply the whole equation by \( x \) to remove the fraction: \( 9x^2 + 1 = 6x \). Move `6x` to the left: \( 9x^2 - 6x + 1 = 0 \). This is the same as \( (3x-1)^2 = 0 \). So, \( 3x-1=0 \), which means \( x=1/3 \). The answer is \( x=1/3 \), and it's a repeated solution.

🎯 Exam Tip: Recognize perfect square trinomials like \( (ax-b)^2 \) or \( (ax+b)^2 \) to quickly factor them and solve for repeated roots.

Question 16. \( \frac{x}{5}+\frac{28}{x+2} = 5 \)
Answer: First, find a common denominator, \( 5(x+2) \), and multiply the entire equation by it to clear fractions. This yields \( x(x+2) + 28 \times 5 = 5 \times 5(x+2) \), which simplifies to \( x^2 + 2x + 140 = 25x + 50 \). Rearrange into standard quadratic form: \( x^2 - 23x + 90 = 0 \). To factor, find two numbers that multiply to 90 and add to -23. These numbers are -18 and -5. Rewrite the middle term: \( x^2 - 18x - 5x + 90 = 0 \). Factor by grouping: \( x(x-18) - 5(x-18) = 0 \), which simplifies to \( (x-18)(x-5) = 0 \). Setting each factor to zero, we get \( x=18 \) or \( x=5 \). Always check that the solutions do not make any original denominators zero.
In simple words: Multiply the whole equation by \( 5(x+2) \) to remove fractions. This gives \( x(x+2) + 140 = 25(x+2) \). Expand and simplify: \( x^2 + 2x + 140 = 25x + 50 \). Bring everything to one side: \( x^2 - 23x + 90 = 0 \). Find two numbers that multiply to 90 and add to -23 (which are -18 and -5). Split the middle term: \( x^2 - 18x - 5x + 90 = 0 \). Group: \( x(x-18) - 5(x-18) = 0 \). So, \( (x-18)(x-5) = 0 \). This means \( x=18 \) or \( x=5 \).

🎯 Exam Tip: Be careful with signs when combining and rearranging terms, especially when moving terms across the equals sign.

Question 17. \( \frac{x}{x-1}+\frac{x-1}{x}=\frac{5}{2} \)
Answer: First, combine the fractions on the left side by finding a common denominator, which is \( x(x-1) \). This gives \( \frac{x^2 + (x-1)^2}{x(x-1)} = \frac{5}{2} \). Expand the numerator and denominator: \( \frac{x^2 + x^2 - 2x + 1}{x^2 - x} = \frac{5}{2} \), which simplifies to \( \frac{2x^2 - 2x + 1}{x^2 - x} = \frac{5}{2} \). Now, cross-multiply: \( 2(2x^2 - 2x + 1) = 5(x^2 - x) \). Expand both sides: \( 4x^2 - 4x + 2 = 5x^2 - 5x \). Rearrange all terms to one side to form a quadratic equation: \( x^2 - x - 2 = 0 \). Factor this equation by finding two numbers that multiply to -2 and add to -1. These numbers are -2 and 1. Rewrite the middle term: \( x^2 - 2x + x - 2 = 0 \). Factor by grouping: \( x(x-2) + 1(x-2) = 0 \), leading to \( (x-2)(x+1) = 0 \). Setting each factor to zero, we find \( x=2 \) or \( x=-1 \). It's crucial that \( x \neq 0 \) and \( x \neq 1 \) in the original equation, which our solutions satisfy.
In simple words: First, join the fractions on the left side: \( \frac{x^2 + (x-1)^2}{x(x-1)} = \frac{5}{2} \). Simplify this to \( \frac{2x^2 - 2x + 1}{x^2 - x} = \frac{5}{2} \). Cross-multiply: \( 2(2x^2 - 2x + 1) = 5(x^2 - x) \). Expand and bring everything to one side: \( x^2 - x - 2 = 0 \). Factor this as \( (x-2)(x+1) = 0 \). So, \( x=2 \) or \( x=-1 \).

🎯 Exam Tip: Always state the restrictions on the variable (values that make the denominator zero) and confirm your solutions are valid.

Question 18. \( \frac{x}{a}-\frac{a+b}{x}=\frac{b(a+b)}{a x} \)
Answer: First, combine the fractions on the left side by finding a common denominator, \( ax \). This gives \( \frac{x^2 - a(a+b)}{ax} = \frac{b(a+b)}{ax} \). Since both sides have \( ax \) in the denominator, we can multiply both sides by \( ax \) (assuming \( a \) and \( x \) are not zero). This leaves \( x^2 - a(a+b) = b(a+b) \). Expand and rearrange all terms to one side: \( x^2 - a^2 - ab = ab + b^2 \), so \( x^2 - a^2 - 2ab - b^2 = 0 \). Recognize the quadratic form of \( (a+b)^2 \): \( x^2 - (a^2 + 2ab + b^2) = 0 \), which is \( x^2 - (a+b)^2 = 0 \). This is a difference of squares, which factors as \( (x - (a+b))(x + (a+b)) = 0 \). Setting each factor to zero, we get \( x = a+b \) or \( x = -(a+b) \). Algebraic manipulation must be done carefully to avoid errors.
In simple words: First, make the left side one fraction: \( \frac{x^2 - a(a+b)}{ax} = \frac{b(a+b)}{ax} \). Since both sides have `ax` below, we can remove it. So, \( x^2 - a(a+b) = b(a+b) \). Expand and move everything to one side: \( x^2 - a^2 - 2ab - b^2 = 0 \). This can be written as \( x^2 - (a+b)^2 = 0 \). Using the difference of squares, \( (x-(a+b))(x+(a+b))=0 \). So, \( x = a+b \) or \( x = -(a+b) \).

🎯 Exam Tip: Watch for common algebraic identities like the difference of squares \( (A^2 - B^2 = (A-B)(A+B)) \) or perfect square trinomials, as they simplify factoring.

Question 19. \( \frac{1}{a+b+x}-\frac{1}{x}=\frac{1}{a}+\frac{1}{b} \)
Answer: First, combine the fractions on both sides of the equation. On the left side, the common denominator is \( x(a+b+x) \), giving \( \frac{x-(a+b+x)}{x(a+b+x)} \). On the right side, the common denominator is \( ab \), giving \( \frac{b+a}{ab} \). This simplifies to \( \frac{-a-b}{x(a+b+x)} = \frac{a+b}{ab} \). We can rewrite \( -a-b \) as \( -(a+b) \), so \( \frac{-(a+b)}{x(a+b+x)} = \frac{a+b}{ab} \). Divide both sides by \( (a+b) \) (assuming \( a+b \neq 0 \)) to get \( \frac{-1}{x(a+b+x)} = \frac{1}{ab} \). Cross-multiply to remove denominators: \( -ab = x(a+b+x) \). Expand and rearrange into standard quadratic form: \( -ab = x^2 + ax + bx \), which becomes \( x^2 + ax + bx + ab = 0 \). Factor by grouping: \( x(x+a) + b(x+a) = 0 \), leading to \( (x+a)(x+b) = 0 \). Setting each factor to zero, we find the solutions \( x=-a \) or \( x=-b \). This problem demonstrates powerful algebraic simplification techniques.
In simple words: First, make each side of the equation a single fraction. The left side becomes \( \frac{x-(a+b+x)}{x(a+b+x)} \), and the right side becomes \( \frac{a+b}{ab} \). This simplifies to \( \frac{-(a+b)}{x(a+b+x)} = \frac{a+b}{ab} \). We can cancel \( (a+b) \) from both sides. Then, cross-multiply: \( -ab = x(a+b+x) \). Expand and bring everything to one side: \( x^2 + ax + bx + ab = 0 \). Factor by grouping: \( x(x+a) + b(x+a) = 0 \). So, \( (x+a)(x+b) = 0 \). This means \( x=-a \) or \( x=-b \).

🎯 Exam Tip: Remember that if you divide by a term like \( (a+b) \), you must consider the case where that term is zero separately, although in this context, it's usually assumed non-zero for the simplification to hold.

Question 20. \( \frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}, x \neq 0, x \neq 2 \)
Answer: First, combine the fractions on the left side using the common denominator \( x(x-2) \). This yields \( \frac{x(x+3) - (1-x)(x-2)}{x(x-2)} = \frac{17}{4} \). Expand and simplify the numerator: \( \frac{x^2+3x - (x-2-x^2+2x)}{x^2-2x} = \frac{17}{4} \), which becomes \( \frac{x^2+3x - (-x^2+3x-2)}{x^2-2x} = \frac{17}{4} \). After simplifying further, we get \( \frac{2x^2 + 2}{x^2 - 2x} = \frac{17}{4} \). Now, cross-multiply: \( 4(2x^2+2) = 17(x^2-2x) \). Expand both sides: \( 8x^2+8 = 17x^2-34x \). Rearrange into standard quadratic form: \( 9x^2 - 34x - 8 = 0 \). To factor this, we need two numbers that multiply to \( (9 \times -8 = -72) \) and add to -34. These numbers are -36 and 2. Rewrite the middle term: \( 9x^2 - 36x + 2x - 8 = 0 \). Factor by grouping: \( 9x(x-4) + 2(x-4) = 0 \), which gives \( (x-4)(9x+2) = 0 \). Setting each factor to zero, we find \( x=4 \) or \( x=-\frac{2}{9} \). Both solutions respect the initial restrictions for \( x \).
In simple words: First, join the fractions on the left side: \( \frac{x(x+3)-(1-x)(x-2)}{x(x-2)} = \frac{17}{4} \). Expand and simplify the top part to get \( \frac{2x^2+2}{x^2-2x} = \frac{17}{4} \). Cross-multiply: \( 4(2x^2+2) = 17(x^2-2x) \). This expands to \( 8x^2+8 = 17x^2-34x \). Bring everything to one side: \( 9x^2 - 34x - 8 = 0 \). Find two numbers that multiply to -72 and add to -34 (which are -36 and 2). Split the middle term: \( 9x^2 - 36x + 2x - 8 = 0 \). Group: \( 9x(x-4) + 2(x-4) = 0 \). So, \( (x-4)(9x+2) = 0 \). This gives \( x=4 \) or \( x=-2/9 \).

🎯 Exam Tip: Expanding and simplifying algebraic expressions in the numerator and denominator correctly is crucial to avoid errors in rational equations.

Question 21. \( \frac{2x}{x-4}+\frac{2x-5}{x-3}=\frac{25}{3} \)
Answer: First, combine the fractions on the left side using the common denominator \( (x-4)(x-3) \). This results in \( \frac{2x(x-3) + (x-4)(2x-5)}{(x-4)(x-3)} = \frac{25}{3} \). Expand the numerator and denominator to simplify: \( \frac{2x^2-6x + 2x^2-5x-8x+20}{x^2-3x-4x+12} = \frac{25}{3} \), which becomes \( \frac{4x^2 - 19x + 20}{x^2 - 7x + 12} = \frac{25}{3} \). Now, cross-multiply: \( 3(4x^2 - 19x + 20) = 25(x^2 - 7x + 12) \). Expand both sides: \( 12x^2 - 57x + 60 = 25x^2 - 175x + 300 \). Rearrange all terms to one side to form a quadratic equation: \( 13x^2 - 118x + 240 = 0 \). To factor this, we need two numbers that multiply to \( (13 \times 240 = 3120) \) and add to -118. These numbers are -78 and -40. Rewrite the middle term: \( 13x^2 - 78x - 40x + 240 = 0 \). Factor by grouping: \( 13x(x-6) - 40(x-6) = 0 \), leading to \( (x-6)(13x-40) = 0 \). Setting each factor to zero, we find \( x=6 \) or \( x=\frac{40}{13} \). Always be careful with arithmetic involving larger numbers.
In simple words: Combine the fractions on the left side: \( \frac{2x(x-3) + (x-4)(2x-5)}{(x-4)(x-3)} = \frac{25}{3} \). Expand and simplify this to \( \frac{4x^2 - 19x + 20}{x^2 - 7x + 12} = \frac{25}{3} \). Cross-multiply: \( 3(4x^2 - 19x + 20) = 25(x^2 - 7x + 12) \). Expand and move all terms to one side: \( 13x^2 - 118x + 240 = 0 \). Find two numbers that multiply to 3120 (from 13x240) and add to -118 (which are -78 and -40). Split the middle term: \( 13x^2 - 78x - 40x + 240 = 0 \). Group: \( 13x(x-6) - 40(x-6) = 0 \). So, \( (x-6)(13x-40) = 0 \). This gives \( x=6 \) or \( x=40/13 \).

🎯 Exam Tip: Solving rational equations often leads to quadratic equations, so a strong grasp of quadratic factoring techniques is essential.

Question 22. \( \frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6}, x \neq 1, -1 \)
Answer: First, combine the fractions on the left side using the common denominator \( (x-1)(x+1) \). This leads to \( \frac{(x+1)^2 - (x-1)^2}{(x-1)(x+1)} = \frac{5}{6} \). Expand the squared terms in the numerator and simplify: \( \frac{(x^2+2x+1) - (x^2-2x+1)}{x^2-1} = \frac{5}{6} \), which simplifies further to \( \frac{4x}{x^2-1} = \frac{5}{6} \). Now, cross-multiply: \( 6(4x) = 5(x^2-1) \). This expands to \( 24x = 5x^2 - 5 \). Rearrange into standard quadratic form: \( 5x^2 - 24x - 5 = 0 \). To factor this, we need two numbers that multiply to \( (5 \times -5 = -25) \) and add to -24. These numbers are -25 and 1. Rewrite the middle term: \( 5x^2 - 25x + x - 5 = 0 \). Factor by grouping: \( 5x(x-5) + 1(x-5) = 0 \), leading to \( (x-5)(5x+1) = 0 \). Setting each factor to zero, we find \( x=5 \) or \( x=-\frac{1}{5} \). These solutions do not conflict with the initial restrictions.
In simple words: Combine the fractions on the left side: \( \frac{(x+1)^2 - (x-1)^2}{(x-1)(x+1)} = \frac{5}{6} \). Simplify the top part to \( 4x \) and the bottom to \( x^2-1 \), so \( \frac{4x}{x^2-1} = \frac{5}{6} \). Cross-multiply: \( 6(4x) = 5(x^2-1) \). This becomes \( 24x = 5x^2 - 5 \). Move everything to one side: \( 5x^2 - 24x - 5 = 0 \). Find two numbers that multiply to -25 and add to -24 (which are -25 and 1). Split the middle term: \( 5x^2 - 25x + x - 5 = 0 \). Group: \( 5x(x-5) + 1(x-5) = 0 \). So, \( (x-5)(5x+1) = 0 \). This gives \( x=5 \) or \( x=-1/5 \).

🎯 Exam Tip: Remember the difference of squares identity \( a^2 - b^2 = (a-b)(a+b) \) and the perfect square identities \( (a \pm b)^2 = a^2 \pm 2ab + b^2 \) when simplifying expressions.

Question 23. \( \sqrt{2}x^2 - 3x - 2\sqrt{2} = 0 \)
Answer: To solve this quadratic equation, we use factorization. We need two numbers that multiply to \( (\sqrt{2} \times -2\sqrt{2} = -4) \) and add up to -3. These numbers are -4 and 1. Rewrite the middle term \( -3x \) as \( -4x + x \): \( \sqrt{2}x^2 - 4x + x - 2\sqrt{2} = 0 \). Factor by grouping. Remember that \( 4 \) can be written as \( 2\sqrt{2} \times \sqrt{2} \). So, \( \sqrt{2}x(x - 2\sqrt{2}) + 1(x - 2\sqrt{2}) = 0 \). This simplifies to \( (x - 2\sqrt{2})(\sqrt{2}x + 1) = 0 \). Setting each factor to zero, we find \( x=2\sqrt{2} \) or \( x=\frac{-1}{\sqrt{2}} \). Equations with radicals require careful factorization.
In simple words: To solve, find two numbers that multiply to -4 (from \( \sqrt{2} \times -2\sqrt{2} \)) and add to -3 (which are -4 and 1). Split the middle term: \( \sqrt{2}x^2 - 4x + x - 2\sqrt{2} = 0 \). Group the terms: \( \sqrt{2}x(x - 2\sqrt{2}) + 1(x - 2\sqrt{2}) = 0 \). So, \( (x - 2\sqrt{2})(\sqrt{2}x + 1) = 0 \). This means \( x=2\sqrt{2} \) or \( x=-1/\sqrt{2} \).

🎯 Exam Tip: When factoring quadratics with irrational coefficients, remember to look for factors that simplify the radical terms during the grouping step.

Question 24. \( a (x^2 + 1) - x (a^2 + 1) = 0 \)
Answer: First, expand the terms in the equation: \( ax^2 + a - a^2x - x = 0 \). Rearrange the terms to prepare for factoring by grouping: \( ax^2 - a^2x - x + a = 0 \). Factor out common terms from each pair: \( ax(x-a) - 1(x-a) = 0 \). This simplifies to \( (x-a)(ax-1) = 0 \). Setting each factor to zero, we find \( x=a \) or \( x=\frac{1}{a} \). This problem showcases factoring quadratics involving literal coefficients.
In simple words: First, open the brackets: \( ax^2 + a - a^2x - x = 0 \). Rearrange the terms: \( ax^2 - a^2x - x + a = 0 \). Group terms and factor out common parts: \( ax(x-a) - 1(x-a) = 0 \). This makes \( (x-a)(ax-1) = 0 \). So, \( x=a \) or \( x=1/a \).

🎯 Exam Tip: When factoring equations with parameters (like 'a' and 'b'), treat the parameters as constants and follow the same factoring steps as with numerical coefficients.

Question 25. Find the solution set of: \( x^2 - 2x - 35 = 0 \)
Answer: To find the solution set for this quadratic equation, we use factorization. We need two numbers that multiply to -35 and add up to -2. These numbers are -7 and 5. We rewrite the middle term \( -2x \) as \( -7x + 5x \): \( x^2 - 7x + 5x - 35 = 0 \). Factor by grouping: \( x(x-7) + 5(x-7) = 0 \), which simplifies to \( (x-7)(x+5) = 0 \). Setting each factor to zero, we find \( x=7 \) or \( x=-5 \). The solution set is therefore \( \{-5, 7\} \). The order of elements in a set does not matter, so \(-5, 7\) is the same as \(7, -5\).
In simple words: To solve, find two numbers that multiply to -35 and add to -2 (which are -7 and 5). Split the middle term: \( x^2 - 7x + 5x - 35 = 0 \). Group the terms: \( x(x-7) + 5(x-7) = 0 \). So, \( (x-7)(x+5) = 0 \). This means \( x=7 \) or \( x=-5 \). The answers are -5 and 7.

🎯 Exam Tip: Always present the solution set using curly braces \( \{\} \) when asked for a "solution set," and list the elements in ascending order for consistency.