Cubic Equations in One Variable
Let \( x_1, x_2, x_3 \) be the roots of the cubic equation in one variable
\[
ax^3 + bx^2 + cx + d = 0
\]
where \( a, b, c, d \in \mathbb{R} \) and \( a \neq 0 \). If one of the roots, say \( x_1 \), is determined, then \( x – x_1 \) must be a linear factor of the polynomial expression \( ax^3 + bx^2 + cx + d \). Thus, by polynomial long division, we can write:
\[
\frac{ax^3 + bx^2 + cx + d}{x – x_1} = Ax^2 + Bx + C
\]
The roots of the remaining quadratic equation
\[
Ax^2 + Bx + C = 0
\]
will yield the other two roots, \( x_2 \) and \( x_3 \), of the original cubic equation.
Example:
Find the solution set of the cubic equation:
\[
x^3 – 8x^2 + 19x – 12 = 0
\]
Since the sum of the coefficients of the given equation is
\[
1 – 8 + 19 – 12 = 0
\]
by the Rational Root Theorem and the factor theorem, \( 1 \) is a root of the equation. Performing synthetic or polynomial long division:
\[
\begin{array}{r|l}
x^3 – 8x^2 + 19x – 12 & x-1 \\
& \rule{25mm}{0.35mm} \\
– \rule{35mm}{0.35mm} & x^2 – 7x + 12 \\
0 &
\end{array}
\]
Now we solve the quotient equation:
\[
x^2 – 7x + 12 = 0 \Rightarrow x = 3 \quad \text{or} \quad x = 4
\]
Thus, the complete solution set is:
\[
S = \{ 1, 3, 4 \}
\]
Example:
Find the solution set of the cubic equation:
\[
x^3 – x^2 – x – 2 = 0
\]
Testing the integer factors of the constant term \(-2\), we find that \(2\) satisfies the equation. Therefore, the first part of our solution set is:
\[
S_1 = \{2\}
\]
Next, dividing the cubic polynomial by \( x – 2 \):
\[
\begin{array}{r|l}
x^3 – x^2 – x – 2 & x-2 \\
& \rule{25mm}{0.35mm} \\
– \rule{35mm}{0.35mm} & x^2 + x + 1 \\
0 &
\end{array}
\]
We find that the remaining quadratic equation has no real solutions since its discriminant is negative:
\[
x^2 + x + 1 = 0 \Rightarrow S_2 = \emptyset
\]
Therefore, the complete real solution set is:
\[
S = S_1 \cup S_2 = \{2\}
\]
Example:
Find the solution set of the cubic equation:
\[
2x^3 – 3x + 1 = 0
\]
Since the sum of the coefficients is
\[
2 – 3 + 1 = 0
\]
we know that \(1\) is a root. Dividing the polynomial by \( x – 1 \):
\[
\begin{array}{r|l}
2x^3 – 3x + 1 & x-1 \\
& \rule{25mm}{0.35mm} \\
– \rule{35mm}{0.35mm} & 2x^2 + 2x – 1 \\
0 &
\end{array}
\]
Applying the quadratic formula to the quotient equation:
\[
2x^2 + 2x – 1 = 0
\]
\[
\Rightarrow x_{2 , 3} = \frac{-2 \pm 2\sqrt{3}}{2 \cdot 2}
\]
Thus, the solution set is:
\[
S = \left\{ \frac{-1 – \sqrt{3}}{2}, \frac{-1 + \sqrt{3}}{2}, 1 \right\}
\]
Vieta’s Formulas for Cubic Equations:
Given a cubic equation with roots \( x_1, x_2, x_3 \):
\[
ax^3 + bx^2 + cx + d = 0
\]
Dividing both sides by the leading coefficient \( a \), we obtain the monic form:
\[
x^3 + \frac{b}{a} x^2 + \frac{c}{a} x + \frac{d}{a} = 0
\]
Alternatively, expanding the factored form of the cubic equation
\[
(x – x_1)(x – x_2)(x – x_3) = 0
\]
and comparing the coefficients yields the following relationships:
Sum of the Roots:
\[
x_1 + x_2 + x_3 = -\frac{b}{a}
\]
Product of the Roots:
\[
x_1 \cdot x_2 \cdot x_3 = -\frac{d}{a}
\]
Sum of the Products of the Roots Taken Two at a Time:
\[
x_1 \cdot x_2 + x_1 \cdot x_3 + x_2 \cdot x_3 = \frac{c}{a}
\]
Example:
The roots of the equation \( x^3 – 4x^2 + 6 = 0 \) are \( x_1, x_2, x_3 \). Find the value of the expression:
\[
A = \frac{x_1}{x_2 x_3} + \frac{x_2}{x_1 x_3} + \frac{x_3}{x_1 x_2}
\]
Finding a common denominator and expanding the numerator using algebraic identities:
\[
A = \frac{x_1}{x_2 x_3} + \frac{x_2}{x_1 x_3} + \frac{x_3}{x_1 x_2} = \frac{x_1^2 + x_2^2 + x_3^2}{x_1 x_2 x_3}
\]
\[
A = \frac{(x_1 + x_2 + x_3)^2 \;- \; 2(x_1 x_2 + x_1 x_3 + x_2 x_3)}{x_1 x_2 x_3}
\]
Substituting the values obtained from Vieta’s formulas:
\[
A = \frac{4^2 – 2 \cdot 1}{-6} = -\frac{7}{3}
\]
QUESTION 34
The roots of the equation
\[
x^3 + (m + 1)x^2 – mx – 2 = 0
\]
are \( x_1, x_2, x_3 \). Given that
\[
\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = -\frac{1}{2}
\]
what is the sum of the roots of this equation?
\[
\text{A) } -2 \quad
\text{B) } -1 \quad
\text{C) } 1 \quad
\text{D) } 2 \quad
\text{E) } 3
\]
Solution:
Finding a common denominator for the given relation:
\[
\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = -\frac{1}{2}
\]
\[
\Rightarrow \frac{x_1 x_2 + x_1 x_3 + x_2 x_3}{x_1 x_2 x_3} = -\frac{1}{2}
\]
Applying Vieta’s formulas:
\[
\Rightarrow \frac{-m}{2} = -\frac{1}{2}
\]
\[
\Rightarrow m = 1
\]
Substituting \( m = 1 \) into the original equation:
\[
x^3 + (m + 1)x^2 – mx – 2 = 0 \Rightarrow x^3 + 2x^2 – x – 2 = 0
\]
Thus, the sum of the roots is:
\[
x_1 + x_2 + x_3 = -2
\]
\(\textbf{Answer: A} \)
QUESTION 35
The cubic equation
\[
x^3 + 3ax + 16 = 0
\]
has a double root (two identical roots). Find the value of \( a \).
\[
\text{A) } -6 \quad
\text{B) } -5 \quad
\text{C) } -4 \quad
\text{D) } 3 \quad
\text{E) } 5
\]
Solution:
Let the roots of the equation be \( x_1, x_2, x_3 \). Since two of the roots are identical, let \( x_1 = x_2 \). Using Vieta’s formulas:
\[
x_1 + x_2 + x_3 = 2x_1 + x_3 = 0
\]
\[
\Rightarrow x_3 = -2x_1
\]
For the product of the roots:
\[
x_1 x_2 x_3 = x_1^2 x_3 = -16
\]
Substituting \( x_3 = -2x_1 \) into the product equation:
\[
\Rightarrow x_1^2 (-2x_1) = -16
\]
\[
\Rightarrow -2x_1^3 = -16
\]
\[
\Rightarrow x_1 = 2
\]
Since \(x_1 = 2\) is a root, it must satisfy the original equation. Substituting \(x = 2\):
\[
x^3 + 3ax + 16 = 0 \Rightarrow 2^3 + 3a \cdot 2 + 16 = 0
\]
\[
\Rightarrow a = -4
\]
\(\textbf{Answer: C} \)
QUESTION 36
The roots of the equation
\[
x^3 + x^2 – 3x + m = 0
\]
are \( x_1, x_2, x_3 \). Given that
\[
x_1^2 \;x_2^2\; x_3 + x_1^2\; x_2 \;x_3^2 + x_1\;x_2^2\;x_3^2 = 3
\]
find the value of \( m \).
\[
\text{A) } -2 \quad
\text{B) } -1 \quad
\text{C) } 0 \quad
\text{D) } 1 \quad
\text{E) } 2
\]
Solution:
Factoring out the greatest common factor \( x_1 x_2 x_3 \):
\[
x_1^2 \;x_2^2\; x_3 + x_1^2\; x_2 \;x_3^2 + x_1\;x_2^2\;x_3^2 = 3
\]
\[\Rightarrow x_1 \;x_2\; x_3 (x_1\;x_2+ x_1\;x_3 + x_2\; x_3)= 3 \]
Using Vieta’s formulas, substitute the expressions for the product and the sum taken two at a time:
\[\Rightarrow -m \cdot (-3)= 3 \Rightarrow m=1 \]
\(\textbf{Answer: D} \)
QUESTION 37
Let \( m \) and \( n \) be non-zero real numbers. Given that the roots of the quadratic equation
\[
x^2 – mx + n = 0
\]
are also roots of the cubic equation
\[
2x^3 – 4mx^2 + 3x – 2mn^2 = 0
\]
find the value of \( n \).
\[
\text{A) } 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]
Solution:
Let \( x_1, x_2 \) be the roots of the quadratic equation, and let \( x_1, x_2, x_3 \) be the roots of the cubic equation. From the quadratic equation, we have:
\[
x_1 + x_2 = m \quad \text{and} \quad x_1 x_2 = n
\]
From the cubic equation, using Vieta’s formulas:
\[
x_1 + x_2 + x_3 = \frac{-(-4m)}{2} = 2m
\]
Substituting \( x_1 + x_2 = m \) into this sum:
\[
m + x_3 = 2m \Rightarrow x_3 = m
\]
Now, using the product of the roots for the cubic equation:
\[
x_1 \;x_2 \; x_3 = \frac{-(-2mn^2)}{2} = mn^2
\]
Substituting \( x_1 x_2 = n \) and \( x_3 = m \):
\[
n \cdot m = mn^2
\]
Since \( m \neq 0 \) and \( n \neq 0 \), dividing both sides by \( mn \) yields:
\[
n = 1
\]
\(\textbf{Answer: A} \)
Vieta’s Formulas for Polynomial Equations of Degree \(n\):
Let \( x_1, x_2, x_3, \dots, x_n \) be the roots of the \(n\)-th degree polynomial equation:
\[
a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 = 0
\]
Sum of the Roots:
\[
x_1 + x_2 + x_3 + \dots + x_n = -\frac{a_{n-1}}{a_n}
\]
Product of the Roots:
\[
x_1 \cdot x_2 \cdot x_3 \cdots x_n = (-1)^n \frac{a_0}{a_n}
\]
Example:
Find the sum and product of the roots of the polynomial equation:
\[
2x^7 – 6x^6 + 3x^2 – x – 4 = 0
\]
Using the generalized formulas for a 7th-degree polynomial (\(n=7\)):
\[
x_1 + x_2 + x_3 + \dots + x_7 = -\frac{-6}{2} = 3
\]
\[
x_1 \cdot x_2 \cdot x_3 \cdots x_7 = (-1)^7 \cdot \frac{-4}{2} = 2
\]
QUESTION 38
All roots of the equation
\[
x^4 + ax^2 + 4 = 0
\]
are real numbers. Which of the following statements must be true regarding these roots?
A) All four roots are positive.
B) Two roots are positive and two roots are negative.
C) Three roots are positive and one root is negative.
D) All four roots are negative.
E) Three roots are negative and one root is positive.
Solution:
Let \( x_1, x_2, x_3, x_4 \) be the roots of the given 4th-degree equation. Since the coefficient of \(x^3\) is zero, the sum of the roots is:
\[
x_1 + x_2 + x_3 + x_4 = 0
\]
The product of the roots is given by:
\[
x_1 \cdot x_2 \cdot x_3 \cdot x_4 = (-1)^4 \cdot \frac{4}{1} = 4
\]
Since the product of the roots is positive, there must be an even number of negative roots. Furthermore, because the equation is symmetric in terms of \(x^2\) (a biquadratic equation), non-zero real roots must occur in pairs of opposites (\(\pm x\)). Since the sum of the roots is zero and the product is positive, the equation must have two distinct positive real roots and two distinct negative real roots.
\(\textbf{Answer: B} \)
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