Answer :
Sure, let's solve each equation step by step using the factorization method.
### 2. [tex]\( a^2 x^2 - b^2 = 0 \)[/tex]
This equation can be solved by recognizing it as a difference of squares. Recall that [tex]\( a^2 - b^2 = (a - b)(a + b) \)[/tex].
[tex]\[ a^2 x^2 - b^2 = 0 \quad \implies \quad (a x - b)(a x + b) = 0 \][/tex]
Setting each factor to zero:
[tex]\[ a x - b = 0 \quad \text{or} \quad a x + b = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ a x = b \quad \implies \quad x = \frac{b}{a} \][/tex]
[tex]\[ a x = -b \quad \implies \quad x = \frac{-b}{a} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{b}{a}, \quad x = \frac{-b}{a} \][/tex]
### 3. [tex]\( b x^2 + a b x = 0 \)[/tex]
This equation can be factored by taking [tex]\( b x \)[/tex] common:
[tex]\[ b x^2 + a b x = 0 \quad \implies \quad b x (x + a) = 0 \][/tex]
Setting each factor to zero:
[tex]\[ b x = 0 \quad \text{or} \quad x + a = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ b x = 0 \quad \implies \quad x = 0 \][/tex]
[tex]\[ x + a = 0 \quad \implies \quad x = -a \][/tex]
So, the solutions are:
[tex]\[ x = 0, \quad x = -a \][/tex]
### 4. [tex]\( x^2 - 2 \sqrt{3} x + 3 = 0 \)[/tex]
This quadratic equation can be solved by recognizing it as a perfect square:
[tex]\[ x^2 - 2 \sqrt{3} x + 3 = 0 \][/tex]
Noting that [tex]\( 3 = (\sqrt{3})^2 \)[/tex], the left-hand side can be written as:
[tex]\[ (x - \sqrt{3})^2 = 0 \][/tex]
Setting the factor to zero:
[tex]\[ x - \sqrt{3} = 0 \quad \implies \quad x = \sqrt{3} \][/tex]
The solution is:
[tex]\[ x = \sqrt{3} \quad (\text{with multiplicity 2}) \][/tex]
### 5. [tex]\( x^2 - \frac{11}{4} x + \frac{15}{8} = 0 \)[/tex]
To solve this, we can use the method of splitting the middle term.
We need two numbers whose product is [tex]\( \frac{15}{8} \)[/tex] and whose sum is [tex]\( -\frac{11}{4} \)[/tex].
Let's split the middle term:
[tex]\[ x^2 - \frac{11}{4} x + \frac{15}{8} = x^2 - \frac{3}{2} x - \frac{5}{4} x + \frac{15}{8} \][/tex]
Factoring by grouping:
[tex]\[ x (x - \frac{3}{2}) - \frac{5}{4} (x - \frac{3}{2}) \][/tex]
[tex]\[ (x - \frac{5}{4})(x - \frac{3}{2}) = 0 \][/tex]
Setting each factor to zero:
[tex]\[ x - \frac{5}{4} = 0 \quad \implies \quad x = \frac{5}{4} \][/tex]
[tex]\[ x - \frac{3}{2} = 0 \quad \implies \quad x = \frac{3}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{5}{4}, \quad x = \frac{3}{2} \][/tex]
### 6. [tex]\( x^2 - \frac{3 a}{2} x - a^2 = 0 \)[/tex]
This can also be solved by the method of splitting the middle term.
We need two numbers whose product is [tex]\(-a^2\)[/tex] and whose sum is [tex]\(-\frac{3a}{2}\)[/tex].
Let's find those numbers and rewrite the equation:
[tex]\[ x^2 - \frac{3a}{2} x - a^2 = x^2 - 2a x + \frac{a}{2} x - a^2 \][/tex]
Factoring by grouping:
[tex]\[ x (x - 2a) + \frac{a}{2}(x - 2a) \][/tex]
[tex]\[ (x - 2a)(x + \frac{a}{2}) = 0 \][/tex]
Setting each factor to zero:
[tex]\[ x - 2a = 0 \quad \implies \quad x = 2a \][/tex]
[tex]\[ x + \frac{a}{2} = 0 \quad \implies \quad x = -\frac{a}{2} \][/tex]
So, the solutions are:
[tex]\[ x = 2a, \quad x = -\frac{a}{2} \][/tex]
### 7. [tex]\( \frac{x+3}{x+2} = \frac{3 x-7}{2 x-3}; x \neq-2, \frac{3}{2} \)[/tex]
To solve this, cross-multiply to eliminate the fractions:
[tex]\[ (x + 3)(2x - 3) = (x + 2)(3x - 7) \][/tex]
Expand both sides:
[tex]\[ 2x^2 - 3x + 6x - 9 = 3x^2 - 7x + 6x - 14 \][/tex]
Combine like terms:
[tex]\[ 2x^2 + 3x - 9 = 3x^2 - x - 14 \][/tex]
Move all terms to one side of the equation:
[tex]\[ 2x^2 + 3x - 9 - 3x^2 + x + 14 = 0 \][/tex]
Combine like terms:
[tex]\[ -x^2 + 4x + 5 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ -x^2 + 4x + 5 = -(x^2 - 4x - 5) = -(x - 5)(x + 1) = 0 \][/tex]
Setting each factor to zero:
[tex]\[ x - 5 = 0 \quad \implies \quad x = 5 \][/tex]
[tex]\[ x + 1 = 0 \quad \implies \quad x = -1 \][/tex]
So, the solutions are:
[tex]\[ x = 5, \quad x = -1 \][/tex]
Both solutions [tex]\(x = 5\)[/tex] and [tex]\(x = -1\)[/tex] do not violate the given condition [tex]\( x \neq -2, \frac{3}{2} \)[/tex]. Thus, the solutions are acceptable.
### 2. [tex]\( a^2 x^2 - b^2 = 0 \)[/tex]
This equation can be solved by recognizing it as a difference of squares. Recall that [tex]\( a^2 - b^2 = (a - b)(a + b) \)[/tex].
[tex]\[ a^2 x^2 - b^2 = 0 \quad \implies \quad (a x - b)(a x + b) = 0 \][/tex]
Setting each factor to zero:
[tex]\[ a x - b = 0 \quad \text{or} \quad a x + b = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ a x = b \quad \implies \quad x = \frac{b}{a} \][/tex]
[tex]\[ a x = -b \quad \implies \quad x = \frac{-b}{a} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{b}{a}, \quad x = \frac{-b}{a} \][/tex]
### 3. [tex]\( b x^2 + a b x = 0 \)[/tex]
This equation can be factored by taking [tex]\( b x \)[/tex] common:
[tex]\[ b x^2 + a b x = 0 \quad \implies \quad b x (x + a) = 0 \][/tex]
Setting each factor to zero:
[tex]\[ b x = 0 \quad \text{or} \quad x + a = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ b x = 0 \quad \implies \quad x = 0 \][/tex]
[tex]\[ x + a = 0 \quad \implies \quad x = -a \][/tex]
So, the solutions are:
[tex]\[ x = 0, \quad x = -a \][/tex]
### 4. [tex]\( x^2 - 2 \sqrt{3} x + 3 = 0 \)[/tex]
This quadratic equation can be solved by recognizing it as a perfect square:
[tex]\[ x^2 - 2 \sqrt{3} x + 3 = 0 \][/tex]
Noting that [tex]\( 3 = (\sqrt{3})^2 \)[/tex], the left-hand side can be written as:
[tex]\[ (x - \sqrt{3})^2 = 0 \][/tex]
Setting the factor to zero:
[tex]\[ x - \sqrt{3} = 0 \quad \implies \quad x = \sqrt{3} \][/tex]
The solution is:
[tex]\[ x = \sqrt{3} \quad (\text{with multiplicity 2}) \][/tex]
### 5. [tex]\( x^2 - \frac{11}{4} x + \frac{15}{8} = 0 \)[/tex]
To solve this, we can use the method of splitting the middle term.
We need two numbers whose product is [tex]\( \frac{15}{8} \)[/tex] and whose sum is [tex]\( -\frac{11}{4} \)[/tex].
Let's split the middle term:
[tex]\[ x^2 - \frac{11}{4} x + \frac{15}{8} = x^2 - \frac{3}{2} x - \frac{5}{4} x + \frac{15}{8} \][/tex]
Factoring by grouping:
[tex]\[ x (x - \frac{3}{2}) - \frac{5}{4} (x - \frac{3}{2}) \][/tex]
[tex]\[ (x - \frac{5}{4})(x - \frac{3}{2}) = 0 \][/tex]
Setting each factor to zero:
[tex]\[ x - \frac{5}{4} = 0 \quad \implies \quad x = \frac{5}{4} \][/tex]
[tex]\[ x - \frac{3}{2} = 0 \quad \implies \quad x = \frac{3}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{5}{4}, \quad x = \frac{3}{2} \][/tex]
### 6. [tex]\( x^2 - \frac{3 a}{2} x - a^2 = 0 \)[/tex]
This can also be solved by the method of splitting the middle term.
We need two numbers whose product is [tex]\(-a^2\)[/tex] and whose sum is [tex]\(-\frac{3a}{2}\)[/tex].
Let's find those numbers and rewrite the equation:
[tex]\[ x^2 - \frac{3a}{2} x - a^2 = x^2 - 2a x + \frac{a}{2} x - a^2 \][/tex]
Factoring by grouping:
[tex]\[ x (x - 2a) + \frac{a}{2}(x - 2a) \][/tex]
[tex]\[ (x - 2a)(x + \frac{a}{2}) = 0 \][/tex]
Setting each factor to zero:
[tex]\[ x - 2a = 0 \quad \implies \quad x = 2a \][/tex]
[tex]\[ x + \frac{a}{2} = 0 \quad \implies \quad x = -\frac{a}{2} \][/tex]
So, the solutions are:
[tex]\[ x = 2a, \quad x = -\frac{a}{2} \][/tex]
### 7. [tex]\( \frac{x+3}{x+2} = \frac{3 x-7}{2 x-3}; x \neq-2, \frac{3}{2} \)[/tex]
To solve this, cross-multiply to eliminate the fractions:
[tex]\[ (x + 3)(2x - 3) = (x + 2)(3x - 7) \][/tex]
Expand both sides:
[tex]\[ 2x^2 - 3x + 6x - 9 = 3x^2 - 7x + 6x - 14 \][/tex]
Combine like terms:
[tex]\[ 2x^2 + 3x - 9 = 3x^2 - x - 14 \][/tex]
Move all terms to one side of the equation:
[tex]\[ 2x^2 + 3x - 9 - 3x^2 + x + 14 = 0 \][/tex]
Combine like terms:
[tex]\[ -x^2 + 4x + 5 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ -x^2 + 4x + 5 = -(x^2 - 4x - 5) = -(x - 5)(x + 1) = 0 \][/tex]
Setting each factor to zero:
[tex]\[ x - 5 = 0 \quad \implies \quad x = 5 \][/tex]
[tex]\[ x + 1 = 0 \quad \implies \quad x = -1 \][/tex]
So, the solutions are:
[tex]\[ x = 5, \quad x = -1 \][/tex]
Both solutions [tex]\(x = 5\)[/tex] and [tex]\(x = -1\)[/tex] do not violate the given condition [tex]\( x \neq -2, \frac{3}{2} \)[/tex]. Thus, the solutions are acceptable.
