Finding a Quadratic Equation Given Its Roots
Let \( x_1, x_2 \) be the roots of the quadratic equation
\[
ax^2 + bx + c = 0
\]
If we express this equation in terms of its linear factors, we can write:
\[
(x – x_1) (x – x_2) = 0
\]
Expanding and rearranging the terms yields:
\[
x^2 – (x_1 + x_2)x + x_1 \cdot x_2 = 0
\]
Warning:
For any non-zero real number \( a \), the equation
\[
x^2 – (x_1 + x_2)x + x_1 \cdot x_2 = 0
\]
is equivalent to the equation
\[
a[x^2 – (x_1 + x_2)x + x_1 \cdot x_2] = 0
\]
This means they share the exact same solution set.
Example:
The equation
\[
x^2 – 2x – 5 = 0
\]
and the equation
\[
4x^2 – 8x – 20 = 0
\]
have the same solution set.
Example:
Let us find the quadratic equation whose solution set is \( \{2 + \sqrt{2}, \quad 2 – \sqrt{2} \} \).
Since the roots are
\[
x_1 = 2 + \sqrt{2}, \quad x_2 = 2 – \sqrt{2}
\]
the sum of the roots is:
\[
x_1 + x_2 = 2 + \sqrt{2} + 2 – \sqrt{2} = 4
\]
The product of the roots is:
\[
x_1 \cdot x_2 = (2 + \sqrt{2}) (2 – \sqrt{2}) = 2
\]
Therefore, using the standard form:
\[
x^2 – (x_1 + x_2)x + x_1 \cdot x_2 = 0
\]
\[
\Rightarrow x^2 – 4x + 2 = 0
\]
Example:
The area of a rectangle is 23 square units, and its perimeter is 20 units. Find the dimensions of this rectangle.
Let \( x_1 \) and \( x_2 \) represent the side lengths of the rectangle.
\[
x_1 \cdot x_2 = 23, \quad 2(x_1 + x_2) = 20 \Rightarrow x_1 + x_2 = 10
\]
Thus, we can set up the quadratic equation:
\[
x^2 – (x_1 + x_2)x + x_1 \cdot x_2 = 0
\]
\[
\Rightarrow x^2 – 10x + 23 = 0
\]
Solving this equation gives the dimensions:
\[
x_1 = 5 + \sqrt{2} \quad \text{units} \quad \text{and} \quad x_2 = 5 – \sqrt{2} \quad \text{units.}
\]
QUESTION 31
Which of the following is the quadratic equation whose roots satisfy the system of equations?
\[
3x_1 + 2x_2 + x_1 x_2 = 2 – x_2
\]
and
\[
x_1 + x_2 = 3 + 2x_1 x_2
\]
\[
\text{A) } x^2-x-1 =0 \quad
\text{B) } x^2+x-1 =0 \quad
\]
\[
\text{C) } x^2-2x+1 =0 \quad
\text{D) } x^2-2x-2=0 \quad
\text{E) } x^2+2x-2=0
\]
Solution:
Rearranging the given equations:
\[
3x_1 + 2x_2 + x_1 x_2 = 2 – x_2 \Rightarrow 3(x_1 + x_2) + x_1 x_2 = 2
\]
\[
x_1 + x_2 = 3 + 2x_1 x_2 \Rightarrow x_1 + x_2 – 2x_1 x_2 = 3
\]
Solving this system simultaneously gives:
\[
x_1 + x_2 = 1 \quad \text{and} \quad x_1 x_2 = -1
\]
Substituting these values into the general form
\[
x^2 – (x_1 + x_2)x + x_1 x_2 = 0
\]
yields:
\[
x^2 – x – 1 = 0
\]
\(\textbf{Answer: A} \)
QUESTION 32
Given the equation \( x^2 – 6x + 2 = 0 \), which of the following is the quadratic equation whose roots are \( x_1 – \frac{1}{x_2} \) and \( x_2 – \frac{1}{x_1} \)?
\[ \text{A) } x^2-3x+11 =0 \quad \quad \text{B) } x^2-6x+1 =0 \]
\[ \text{C) } 2x^2-6x+1 =0 \quad \quad \text{D) } 2x^2+6x+1=0 \]
\[ \text{E) } 2x^2+6x-1=0 \]
Solution:
Let the roots of the target equation be \( a \) and \( b \)
\[ a = x_1 – \frac{1}{x_2}, \quad b = x_2 – \frac{1}{x_1} \]
First, let’s find the sum of the new roots, \( a + b \):
\[ a + b = x_1 – \frac{1}{x_2} + x_2 – \frac{1}{x_1} \]
\[
= x_1 + x_2 – \frac{x_1 + x_2}{x_1 x_2} = 6 – \frac{6}{2} = 3
\]
Next, let’s find the product of the new roots, \( a \cdot b \):
\[ a \cdot b = \left( x_1 – \frac{1}{x_2} \right) \left( x_2 – \frac{1}{x_1} \right) \]
\[
= x_1 x_2 – 1 – 1 + \frac{1}{x_1 x_2}
\]
\[
= 2 – 2 + \frac{1}{2} = \frac{1}{2}
\]
The target equation can be written as:
\[ x^2 – (a + b)x + ab = 0 \]
\[
x^2 – 3x + \frac{1}{2} = 0
\]
Multiplying by 2 to clear the fraction gives:
\[
\Rightarrow 2x^2 – 6x + 1 = 0
\]
\(\textbf{Answer: C} \)
QUESTION 33
Which of the following represents the quadratic equation whose roots are the reciprocals of the roots of \( x^2 + 2x – 5 = 0 \)
\[ \text{A) } x^2+5x-1 =0 \quad \quad \text{B) } x^2-5x+1 =0 \]
\[ \text{C) } 5x^2+3x-1 =0 \quad \quad \text{D) } 5x^2-2x-1=0 \]
\[ \text{E) } 5x^2+2x-1=0 \]
Solution:
Let \( x_1 \) be a root of the given equation \( x^2 + 2x – 5 = 0 \), and let \( a \) be a root of the required equation. Therefore:
\[
a = \frac{1}{x_1} \Rightarrow x_1 = \frac{1}{a}
\]
Substituting \( x_1 = \frac{1}{a} \) into the original equation:
\[
x^2 + 2x – 5 = 0 \Rightarrow \left( \frac{1}{a} \right)^2 + 2 \cdot \left( \frac{1}{a} \right) – 5 = 0
\]
\[
\Rightarrow 1 + 2a – 5a^2 = 0
\]
\[
\Rightarrow 5a^2 – 2a – 1 = 0
\]
Expressing this in terms of the standard variable \( x \):
\[
\Rightarrow 5x^2 – 2x – 1 = 0
\]
\(\textbf{Answer: D} \)
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