Equations Reducible to Quadratic Equations
Some higher-degree or non-linear equations can be transformed into a standard quadratic equation in one variable by performing appropriate algebraic manipulations or substitutions.
1) Equations Involving the Product or Quotient of Polynomials:
An $n$-th degree polynomial equation in one variable has exactly $n$ roots (which may be real or complex).
Example:
Find the solution set of the equation $x^4 – 3x^3 + 2x^2 – 6x = 0$.
\[
x^4 – 3x^3 + 2x^2 – 6x = 0 \Rightarrow x(x^3 – 3x^2 + 2x – 6) = 0
\]
\[
\Rightarrow x\left[x^2(x – 3) + 2(x – 3)\right] = 0
\]
\[
\Rightarrow x(x – 3)(x^2 + 2) = 0
\]
\[
\Rightarrow x = 0 \quad \text{or} \quad x – 3 = 0 \quad (\text{since } x^2 + 2 \neq 0 \text{ for real numbers})
\]
\[
\Rightarrow S = \{ 0, 3 \}
\]
Example:
Find the solution set of the equation $(2x – 4)(x^2 – 2x – 2) = 0$.
\[
(2x – 4)(x^2 – 2x – 2) = 0
\]
\[
\Rightarrow 2x – 4 = 0 \quad \text{or} \quad x^2 – 2x – 2 = 0
\]
Using the quadratic formula for the second part:
\[
x_1 = 2 \quad \text{or} \quad x_2 = 1 + \sqrt{3} \quad \text{or} \quad x_3 = 1 – \sqrt{3}
\]
\[
\Rightarrow S = \{ 1 – \sqrt{3}, 2, 1 + \sqrt{3} \}
\]
Example:
Find the solution set of the equation $ \displaystyle \frac{x^3 – 2x^2 – x + 2}{x^2 – x – 2} = 0$.
\[
\frac{x^3 – 2x^2 – x + 2}{x^2 – x – 2} = 0 \Rightarrow \frac{x^2(x – 2) – (x – 2)}{(x – 2)(x + 1)} = 0
\]
\[
\Rightarrow \frac{(x – 2)(x^2 – 1)}{(x – 2)(x + 1)} = 0
\]
From here, we find the roots of the numerator while ensuring the denominator is not zero:
\[
(x – 2)(x^2 – 1) = 0 \quad \text{and} \quad (x – 2)(x + 1) \neq 0
\]
\[
\Rightarrow (x = 2 \text{ or } x = \pm 1) \quad \text{and} \quad (x \neq 2 \text{ and } x \neq -1)
\]
Thus, $x = 2$ and $x = -1$ are excluded as extraneous solutions (they make the denominator zero).
\[
\Rightarrow S = \{ 1 \}
\]
QUESTION 4
Let $m$ be a non-zero real number.
\[
\frac{mx^3 + (2 + m)x^2 – x – 3}{x – 1} = 0
\]
Given that this equation has exactly two distinct real roots, what is the sum of these roots?
\[
\text{A) } 0 \quad
\text{B) } -1 \quad
\text{C) } -2 \quad
\text{D) } -3 \quad
\text{E) } -4
\]
Solution:
Since the numerator is a third-degree polynomial, it has three roots. For the rational equation to yield exactly two real roots, one of the numerator’s roots must match the root of the denominator, thereby canceling out.
The root of the denominator is:
\[
x – 1 = 0 \Rightarrow x = 1
\]
Substituting $x = 1$ into the numerator to make it zero:
\[
m(1)^3 + (2 + m)(1)^2 – 1 – 3 = 0
\]
\[
m + 2 + m – 1 – 3 = 0 \Rightarrow 2m – 2 = 0 \Rightarrow m = 1
\]
Now rewrite and factor the equation with $m = 1$:
\[
\frac{x^3 + 3x^2 – x – 3}{x – 1} = 0
\]
\[
\Rightarrow \frac{x^2(x + 3) – 1(x + 3)}{x – 1} = 0 \Rightarrow \frac{(x + 3)(x^2 – 1)}{x – 1} = 0
\]
\[
\Rightarrow \frac{(x + 3)(x – 1)(x + 1)}{x – 1} = 0
\]
Since $x – 1 \neq 0$, we can cancel the common factor:
\[
(x + 3)(x + 1) = 0 \quad \text{for } x \neq 1
\]
\[
\Rightarrow x_1 = -3 \quad \text{and} \quad x_2 = -1
\]
The sum of these two real roots is:
\[
x_1 + x_2 = -3 + (-1) = -4
\]
\(\textbf{Correct Answer: E} \)
QUESTION 5
What is the sum of the roots of the following equation?
\[
\frac{x – \frac{9}{x}}{x^2 – x + 1 – \frac{1}{x}} + \frac{1}{x – 1} = 0
\]
\[
\text{A) } -1 \quad
\text{B) } 0 \quad
\text{C) } 1 \quad
\text{D) } 2 \quad
\text{E) } 3
\]
Solution:
First, find a common denominator for both the numerator and denominator of the main fraction:
\[
\frac{\frac{x^2 – 9}{x}}{\frac{x(x^2 – x + 1) – 1}{x}} + \frac{1}{x – 1} = 0
\]
\[
\Rightarrow \frac{\frac{x^2 – 9}{x}}{\frac{x^3 – x^2 + x – 1}{x}} + \frac{1}{x – 1} = 0
\]
Simplify by canceling out the $x$ in the denominators:
\[
\frac{x^2 – 9}{x^3 – x^2 + x – 1} + \frac{1}{x – 1} = 0
\]
Factor the denominator by grouping: $x^3 – x^2 + x – 1 = x^2(x – 1) + 1(x – 1) = (x – 1)(x^2 + 1)$.
\[
\Rightarrow \frac{x^2 – 9}{(x – 1)(x^2 + 1)} + \frac{1}{x – 1} = 0
\]
Combine the fractions using the common denominator $(x – 1)(x^2 + 1)$:
\[
\Rightarrow \frac{x^2 – 9 + 1(x^2 + 1)}{(x – 1)(x^2 + 1)} = 0
\]
\[
\Rightarrow \frac{2x^2 – 8}{(x – 1)(x^2 + 1)} = 0 \Rightarrow \frac{2(x^2 – 4)}{(x – 1)(x^2 + 1)} = 0
\]
For this expression to equal zero, the numerator must be zero, and the denominator must not equal zero:
\[
2(x^2 – 4) = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2
\]
Since neither $x = 2$ nor $x = -2$ makes the denominator $(x – 1)(x^2 + 1)$ zero, both are valid solutions.
The sum of the roots is:
\[
x_1 + x_2 = -2 + 2 = 0
\]
\(\textbf{Correct Answer: B} \)
QUESTION 6
What is the solution set of the following equation over the real numbers?
\[
\frac{1}{x – 2} + \frac{4 – 8x}{x^3 – 8} = \frac{1}{x^2 + 2x + 4}
\]
\[
\text{A) } \{3\} \quad
\text{B) } \{1, 3\} \quad
\text{C) } \{2\} \quad
\text{D) } \{2, 5\} \quad
\text{E) } \{5\}
\]
Solution:
Note that $x^3 – 8 = (x – 2)(x^2 + 2x + 4)$ by the difference of cubes identity. Bring all terms to one side over a common denominator:
\[
\frac{1}{x – 2} + \frac{4 – 8x}{(x – 2)(x^2 + 2x + 4)} – \frac{1}{x^2 + 2x + 4} = 0
\]
Equate the denominators:
\[
\frac{(x^2 + 2x + 4) + (4 – 8x) – (x – 2)}{(x – 2)(x^2 + 2x + 4)} = 0
\]
\[
\Rightarrow \frac{x^2 – 7x + 10}{x^3 – 8} = 0
\]
Factor the numerator:
\[
\Rightarrow \frac{(x – 2)(x – 5)}{x^3 – 8} = 0
\]
Setting the numerator to zero gives $x = 2$ or $x = 5$.
However, we must satisfy the domain constraint: $x^3 – 8 \neq 0 \Rightarrow x \neq 2$.
Therefore, $x = 2$ is an extraneous solution.
The only valid solution is:
\[
S = \{ 5 \}
\]
\(\textbf{Correct Answer: E} \)
QUESTION 7
What is the product of the roots of the equation?
\[
\frac{x + 1}{x – 2} – \frac{1}{x + 3} = \frac{9}{x^2 + x – 6}
\]
\[
\text{A) } -2 \quad
\text{B) } -3 \quad
\text{C) } -4 \quad
\text{D) } -5 \quad
\text{E) } -6
\]
Solution:
Factor the denominator on the right side: $x^2 + x – 6 = (x – 2)(x + 3)$.
Move all terms to the left side and combine under the common denominator:
\[
\frac{(x + 1)(x + 3) – 1(x – 2) – 9}{(x – 2)(x + 3)} = 0
\]
\[
\Rightarrow \frac{(x^2 + 4x + 3) – x + 2 – 9}{(x – 2)(x + 3)} = 0
\]
\[
\Rightarrow \frac{x^2 + 3x – 4}{(x – 2)(x + 3)} = 0
\]
Factor the new numerator:
\[
\frac{(x – 1)(x + 4)}{(x – 2)(x + 3)} = 0
\]
This yields candidate roots of $x = 1$ and $x = -4$. Neither value makes the denominator zero ($x \neq 2, x \neq -3$).
Thus, the product of the roots is:
\[
x_1 \cdot x_2 = 1 \cdot (-4) = -4
\]
\(\textbf{Correct Answer: C} \)
2)Equations Solvable Using Substitution (Auxiliary Variables):
Example:
Find the solution set of the equation $x^6 – 7x^3 – 8 = 0$.
Let $x^3 = t$. Substituting this into the equation transforms it into a quadratic:
\[
t^2 – 7t – 8 = 0 \Rightarrow (t – 8)(t + 1) = 0
\]
Substitute back $t = x^3$:
\[
(x^3 – 8)(x^3 + 1) = 0
\]
\[
\Rightarrow (x – 2)(x^2 + 2x + 4)(x + 1)(x^2 – x + 1) = 0
\]
For real roots, the quadratic terms $x^2 + 2x + 4 = 0$ and $x^2 – x + 1 = 0$ have a negative discriminant ($\Delta < 0$) and yield no real solutions.
Hence, the only real solutions come from:
\[
x – 2 = 0 \Rightarrow x_1 = 2 \quad \text{or} \quad x + 1 = 0 \Rightarrow x_2 = -1
\]
The real solution set is:
\[
S = \{ -1, 2 \}
\]
Example:
Find the solution set of the equation $2x + \sqrt{x} – 6 = 0$.
Let $\sqrt{x} = t$ (where $t \geq 0$). Then $x = t^2$:
\[
2t^2 + t – 6 = 0 \Rightarrow (2t – 3)(t + 2) = 0
\]
\[
\Rightarrow t = \frac{3}{2} \quad \text{or} \quad t = -2
\]
Substitute back $t = \sqrt{x}$:
\[
\sqrt{x} = \frac{3}{2} \Rightarrow x = \frac{9}{4}
\]
\[
\sqrt{x} = -2 \Rightarrow \text{No real solution (the principal square root cannot be negative)}
\]
The final solution set is:
\[
S = \left\{ \frac{9}{4} \right\}
\]
QUESTION 8
Which of the following is one of the roots of the equation?
\[ 4^{(x^2)} – 15 \cdot 2^{(x^2)} – 16 = 0 \]
\[
\text{A) } 2 \quad
\text{B) } 3 \quad
\text{C) } 4 \quad
\text{D) } 5 \quad
\text{E) } 6
\]
Solution:
Rewrite the first term as $4^{(x^2)} = (2^2)^{(x^2)} = (2^{(x^2)})^2$.
Let $2^{(x^2)} = t$ (where $t > 0$). The equation becomes:
\[
t^2 – 15t – 16 = 0 \Rightarrow (t – 16)(t + 1) = 0
\]
\[
\Rightarrow t = 16 \quad \text{or} \quad t = -1
\]
Since $t = 2^{(x^2)}$ must be strictly positive, $t = -1$ is impossible.
Set $t = 16$:
\[
2^{(x^2)} = 16 \Rightarrow 2^{(x^2)} = 2^4
\]
\[
\Rightarrow x^2 = 4 \Rightarrow x = 2 \quad \text{or} \quad x = -2
\]
Option A contains the root $2$.
\(\textbf{Correct Answer: A} \)
QUESTION 9
What is the solution set of the equation ?
\[\sqrt[5]{x^4} + \sqrt[5]{x^2} – 2 = 0 \]
\[
\text{A) } \{1, 2\} \quad
\text{B) } \{-1, 1\} \quad
\text{C) } \{3\} \quad
\text{D) } \{-3\} \quad
\text{E) } \{-2, 2\}
\]
Solution:
Notice that $\sqrt[5]{x^4} = (\sqrt[5]{x^2})^2$. Let $\sqrt[5]{x^2} = t$:
\[
t^2 + t – 2 = 0 \Rightarrow (t + 2)(t – 1) = 0
\]
\[
\Rightarrow t = 1 \quad \text{or} \quad t = -2
\]
Substitute back $t = \sqrt[5]{x^2}$:
1. $\sqrt[5]{x^2} = 1 \Rightarrow x^2 = 1^5 = 1 \Rightarrow x = \pm 1$
2. $\sqrt[5]{x^2} = -2 \Rightarrow x^2 = (-2)^5 = -32$, which yields no real solutions since a squared real number cannot be negative.
The real solution set is:
\[
S = \{-1, 1\}
\]
\(\textbf{Correct Answer: B} \)
QUESTION 10
What value of $x$ satisfies the equation ?
\[\sqrt[3]{x – 7} + \frac{1}{\sqrt[3]{x – 7}} = 2 \]
\[
\text{A) } 4 \quad
\text{B) } 5 \quad
\text{C) } 6 \quad
\text{D) } 7 \quad
\text{E) } 8
\]
Solution:
Let $\sqrt[3]{x – 7} = t$:
\[
t + \frac{1}{t} = 2
\]
Multiply both sides by $t$:
\[
t^2 + 1 = 2t \Rightarrow t^2 – 2t + 1 = 0
\]
\[
\Rightarrow (t – 1)^2 = 0 \Rightarrow t = 1
\]
Substitute back to solve for $x$:
\[
\sqrt[3]{x – 7} = 1 \Rightarrow x – 7 = 1^3 \Rightarrow x = 8
\]
\(\textbf{Correct Answer: E} \)
QUESTION 11
What is one of the real roots of the equation ?
\[ x^2 + \frac{1}{x^2} – 3x – \frac{3}{x} + 4 = 0 \]
\[
\text{A) } -2 \quad
\text{B) } -1 \quad
\text{C) } 1 \quad
\text{D) } 2 \quad
\text{E) } 3
\]
Solution:
Group related terms and complete the square for the first pair:
\[
\left(x^2 + \frac{1}{x^2}\right) – 3\left(x + \frac{1}{x}\right) + 4 = 0
\]
Since $\left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}$, we can replace $x^2 + \frac{1}{x^2}$ with $\left(x + \frac{1}{x}\right)^2 – 2$:
\[
\left[\left(x + \frac{1}{x}\right)^2 – 2\right] – 3\left(x + \frac{1}{x}\right) + 4 = 0
\]
\[
\Rightarrow \left(x + \frac{1}{x}\right)^2 – 3\left(x + \frac{1}{x}\right) + 2 = 0
\]
Let $x + \frac{1}{x} = t$:
\[
t^2 – 3t + 2 = 0 \Rightarrow (t – 1)(t – 2) = 0 \Rightarrow t = 1 \text{ or } t = 2
\]
Now, substitute back to find $x$:
1. $x + \frac{1}{x} = 1 \Rightarrow x^2 – x + 1 = 0$. The discriminant is $\Delta = (-1)^2 – 4(1)(1) = -3 < 0$, so there are no real roots.
2. $x + \frac{1}{x} = 2 \Rightarrow x^2 – 2x + 1 = 0 \Rightarrow (x – 1)^2 = 0 \Rightarrow x = 1$.
The single real root is $1$.
\(\textbf{Correct Answer: C} \)
3) Radical (Irrational) Equations:
An equation containing a radical profile like:
\[
\sqrt[2n]{f(x)} = g(x)
\]
can be solved by raising both sides to the $2n$-th power to eliminate the radical. This produces a polynomial equation that can be solved normally.
CRITICAL: Raising an equation to an even power can introduce extraneous solutions (false roots). Therefore, it is mandatory to test all final candidate solutions back in the original equation.
Example:
Find the solution set of the equation.
\[\sqrt{2x + 3} – 2x = 1 \]
Isolate the radical:
\[
\sqrt{2x + 3} = 1 + 2x
\]
Square both sides:
\[
(\sqrt{2x + 3})^2 = (1 + 2x)^2 \Rightarrow 2x + 3 = 1 + 4x + 4x^2
\]
\[
\Rightarrow 4x^2 + 2x – 2 = 0 \Rightarrow 2x^2 + x – 1 = 0
\]
\[
\Rightarrow (2x – 1)(x + 1) = 0 \Rightarrow x_1 = \frac{1}{2}, \quad x_2 = -1
\]
Check candidates in the original expression:
– For $x = \frac{1}{2}$: $\sqrt{2(\frac{1}{2}) + 3} – 2(\frac{1}{2}) = \sqrt{4} – 1 = 2 – 1 = 1$ (Valid)
– For $x = -1$: $\sqrt{2(-1) + 3} – 2(-1) = \sqrt{1} + 2 = 1 + 2 = 3 \neq 1$ (Extraneous)
\[
\Rightarrow S = \left\{ \frac{1}{2} \right\}
\]
QUESTION 12
What value of $x$ satisfies the equation ?
\[ 2x + \sqrt{4x^2 – x – 5} = 7 \]
\[
\text{A) } 5 \quad
\text{B) } 4 \quad
\text{C) } 3 \quad
\text{D) } 2 \quad
\text{E) } 1
\]
Solution:
Isolate the radical and square both sides:
\[
\sqrt{4x^2 – x – 5} = 7 – 2x
\]
\[
(\sqrt{4x^2 – x – 5})^2 = (7 – 2x)^2
\]
\[
4x^2 – x – 5 = 49 – 28x + 4x^2
\]
Cancel out $4x^2$ from both sides:
\[
-x – 5 = 49 – 28x
\]
\[
27x = 54 \Rightarrow x = 2
\]
Verify the solution:
$2(2) + \sqrt{4(2)^2 – 2 – 5} = 4 + \sqrt{16 – 7} = 4 + \sqrt{9} = 4 + 3 = 7$ (Valid)
\(\textbf{Correct Answer: D} \)
QUESTION 13
What is the sum of the roots of the equation ?
\[ \sqrt{2x + 3} – \sqrt{x – 2} = 2 \]
\[
\text{A) } 16 \quad
\text{B) } 14 \quad
\text{C) } 12 \quad
\text{D) } 10 \quad
\text{E) } 8
\]
Solution:
Isolate one radical:
\[
\sqrt{2x + 3} = 2 + \sqrt{x – 2}
\]
Square both sides:
\[
2x + 3 = 4 + 4\sqrt{x – 2} + (x – 2)
\]
\[
2x + 3 = x + 2 + 4\sqrt{x – 2}
\]
Isolate the remaining radical:
\[
x + 1 = 4\sqrt{x – 2}
\]
Square both sides again:
\[
(x + 1)^2 = (4\sqrt{x – 2})^2
\]
\[
x^2 + 2x + 1 = 16(x – 2) \Rightarrow x^2 + 2x + 1 = 16x – 32
\]
\[
x^2 – 14x + 33 = 0 \Rightarrow (x – 3)(x – 11) = 0 \Rightarrow x_1 = 3, \quad x_2 = 11
\]
Verify both candidates in the original equation:
– For $x = 3$: $\sqrt{2(3)+3} – \sqrt{3-2} = \sqrt{9} – \sqrt{1} = 3 – 1 = 2$ (Valid)
– For $x = 11$: $\sqrt{2(11)+3} – \sqrt{11-2} = \sqrt{25} – \sqrt{9} = 5 – 3 = 2$ (Valid)
Both are valid solutions. Their sum is:
\[
x_1 + x_2 = 3 + 11 = 14
\]
\(\textbf{Correct Answer: B} \)
QUESTION 14
Which of the following is the solution set of the equation ?
\[\sqrt[3]{x^2 + 4x – 5} = x – 1 \]
\[
\text{A) } \{-1, 2, 4\} \quad
\text{B) } \{1, 3, 4\} \quad
\text{C) } \{1, 2, 3\} \quad
\text{D) } \{-1, 1, 4\} \quad
\text{E) } \{-1, 2, 3\}
\]
Solution:
Cube both sides to eliminate the radical:
\[
(\sqrt[3]{x^2 + 4x – 5})^3 = (x – 1)^3
\]
\[
x^2 + 4x – 5 = x^3 – 3x^2 + 3x – 1
\]
Rearrange terms into a single polynomial equation:
\[
x^3 – 4x^2 – x + 4 = 0
\]
Factor by grouping:
\[
x^2(x – 4) – 1(x – 4) = 0 \Rightarrow (x – 4)(x^2 – 1) = 0
\]
\[
\Rightarrow (x – 4)(x – 1)(x + 1) = 0 \Rightarrow x = 4, \quad x = 1, \quad x = -1
\]
Since the index of the radical is odd ($3$), the domain is all real numbers and no extraneous roots are created by cubing. All three values are valid.
\[
S = \{-1, 1, 4\}
\]
\(\textbf{Correct Answer: D} \)
QUESTION 15
Which of the following is the solution set of the equation ?
\[\sqrt[4]{2x^2 + x + 6} = \sqrt{x + 2} \]
\[
\text{A) } \{-1, 3\} \quad
\text{B) } \{-1, 2\} \quad
\text{C) } \{1, 2\} \quad
\text{D) } \{1, 3\} \quad
\text{E) } \{2, 3\}
\]
Solution:
Raise both sides to the 4th power:
\[
(\sqrt[4]{2x^2 + x + 6})^4 = (\sqrt{x + 2})^4
\]
\[
2x^2 + x + 6 = (x + 2)^2 \Rightarrow 2x^2 + x + 6 = x^2 + 4x + 4
\]
\[
x^2 – 3x + 2 = 0 \Rightarrow (x – 1)(x – 2) = 0 \Rightarrow x_1 = 1, \quad x_2 = 2
\]
Since both sides are roots of even degree, we must double-check that the radicands remain non-negative ($\geq 0$). Both $x = 1$ and $x = 2$ yield positive values inside the roots, satisfying the equation seamlessly.
\[
S = \{1, 2\}
\]
\(\textbf{Correct Answer: C} \)
QUESTION 16
Which of the following is the solution set of the equation ?
\[\sqrt[3]{3x – 1} = \sqrt{x + 1} \]
\[
\text{A) } \{1, 2, 3\} \quad
\text{B) } \{0, 2\} \quad
\text{C) } \{1, 3\} \quad
\text{D) } \{0, 3\} \quad
\text{E) } \{3\}
\]
Solution:
To clear both the cube root (index 3) and square root (index 2), raise both sides to their least common multiple power, which is the 6th power:
\[
(\sqrt[3]{3x – 1})^6 = (\sqrt{x + 1})^6
\]
\[
(3x – 1)^2 = (x + 1)^3
\]
Expand both sides:
\[
9x^2 – 6x + 1 = x^3 + 3x^2 + 3x + 1
\]
Move all terms to one side:
\[
x^3 – 6x^2 + 9x = 0
\]
Factor the polynomial:
\[
x(x^2 – 6x + 9) = 0 \Rightarrow x(x – 3)^2 = 0 \Rightarrow x = 0 \quad \text{or} \quad x = 3
\]
Test the solutions in the original equation:
– For $x = 0$: $\sqrt[3]{3(0)-1} = \sqrt[3]{-1} = -1$, whereas $\sqrt{0+1} = 1$. Since $-1 \neq 1$, $x = 0$ is extraneous.
– For $x = 3$: $\sqrt[3]{3(3)-1} = \sqrt[3]{8} = 2$, and $\sqrt{3+1} = \sqrt{4} = 2$. (Valid)
\[
S = \{3\}
\]
\(\textbf{Correct Answer: E} \)
QUESTION 17
What is the sum of the real roots of the equation ?
\[ x^2 + x + \sqrt{x^2 + x + 1} = 1 \]
\[
\text{A) } -1 \quad
\text{B) } 0 \quad
\text{C) } 1 \quad
\text{D) } 2 \quad
\text{E) } 3
\]
Solution:
Add 1 to both sides of the equation to match the structure under the radical:
\[
x^2 + x + 1 + \sqrt{x^2 + x + 1} = 2
\]
Let $\sqrt{x^2 + x + 1} = t$ (where $t \geq 0$). The equation turns into:
\[
t^2 + t = 2 \Rightarrow t^2 + t – 2 = 0
\]
\[
(t + 2)(t – 1) = 0 \Rightarrow t = -2 \quad \text{or} \quad t = 1
\]
Since $t \geq 0$, reject $t = -2$. Solve for $t = 1$:
\[
\sqrt{x^2 + x + 1} = 1 \Rightarrow x^2 + x + 1 = 1
\]
\[
x^2 + x = 0 \Rightarrow x(x + 1) = 0 \Rightarrow x_1 = 0 \quad \text{and} \quad x_2 = -1
\]
Both values satisfy the original equation. Their sum is:
\[
0 + (-1) = -1
\]
\(\textbf{Correct Answer: A} \)
4) Absolute Value Equations:
To solve equations containing absolute values, we break the solution down into cases based on the critical points that make the expression inside the absolute value zero.
Find the solution set of the equation $x^2 – |x – 3| + 1 = 0$.
The critical point is $x = 3$.
Case 1: $x – 3 \geq 0 \Rightarrow x \geq 3$
In this domain, $|x – 3| = x – 3$. The equation becomes:
\[
x^2 – (x – 3) + 1 = 0 \Rightarrow x^2 – x + 4 = 0
\]
The discriminant is $\Delta = (-1)^2 – 4(1)(4) = -15 < 0$, which yields no real roots ($S_1 = \emptyset$).
Case 2: $x – 3 < 0 \Rightarrow x < 3$
In this domain, $|x – 3| = -(x – 3) = -x + 3$. The equation becomes:
\[
x^2 – [-(x – 3)] + 1 = 0 \Rightarrow x^2 + x – 3 + 1 = 0
\]
\[
x^2 + x – 2 = 0 \Rightarrow (x + 2)(x – 1) = 0 \Rightarrow x_1 = -2, \quad x_2 = 1
\]
Since both $x = -2$ and $x = 1$ satisfy the condition $x < 3$, both are valid ($S_2 = \{-2, 1\}$).
The complete solution set is:
\[
S = S_1 \cup S_2 = \{-2, 1\}
\]
QUESTION 18
What is the solution set of the equation $2x^2 – x|x – 2| + 1 = 0$ over the real numbers?
\[
\text{A) } \{3\} \quad
\text{B) } \{1\} \quad
\text{C) } \{2\} \quad
\text{D) } \{-1\} \quad
\text{E) } \emptyset
\]
Solution:
Case 1: $x \geq 2 \Rightarrow |x – 2| = x – 2$
\[
2x^2 – x(x – 2) + 1 = 0 \Rightarrow 2x^2 – x^2 + 2x + 1 = 0
\]
\[
x^2 + 2x + 1 = 0 \Rightarrow (x + 1)^2 = 0 \Rightarrow x = -1
\]
However, $x = -1$ does not satisfy our domain condition $x \geq 2$. Thus, this case yields no solutions ($S_1 = \emptyset$).
Case 2: $x < 2 \Rightarrow |x – 2| = -(x – 2)$
\[
2x^2 – x[-(x – 2)] + 1 = 0 \Rightarrow 2x^2 + x^2 – 2x + 1 = 0
\]
\[
3x^2 – 2x + 1 = 0
\]
The discriminant is $\Delta = (-2)^2 – 4(3)(1) = 4 – 12 = -8 < 0$. No real roots exist here ($S_2 = \emptyset$).
Combining both cases:
\[
S = S_1 \cup S_2 = \emptyset
\]
\(\textbf{Correct Answer: E} \)
QUESTION 19
Which of the following is the solution set of the equation ?
\[|x^2 + x – 1| = |2x + 1| \]
\[
\text{A) } \{-2, -1, 0\} \quad
\text{B) } \{-1, 0, 2, 3\} \quad
\text{C) } \{-3, 0, 2, 3\} \quad
\text{D) } \{0, 1, 2\} \quad
\text{E) } \{-3, -1, 0, 2\}
\]
Solution:
An equation of the form $|A| = |B|$ can be split directly into two branches: $A = B$ or $A = -B$.
Branch 1:
\[
x^2 + x – 1 = 2x + 1 \Rightarrow x^2 – x – 2 = 0
\]
\[
(x – 2)(x + 1) = 0 \Rightarrow x_1 = 2, \quad x_2 = -1
\]
Branch 2:
\[
x^2 + x – 1 = -(2x + 1) \Rightarrow x^2 + x – 1 = -2x – 1
\]
\[
x^2 + 3x = 0 \Rightarrow x(x + 3) = 0 \Rightarrow x_3 = 0, \quad x_4 = -3
\]
Collecting all roots gives:
\[
S = \{-3, -1, 0, 2\}
\]
\(\textbf{Correct Answer: E} \)
QUESTION 20
Which of the following is the solution set of the equation ?
\[ |2x – 1| = |x + 1| \]
\[
\text{A) } \{0, 2\} \quad
\text{B) } \{-1, 0\} \quad
\text{C) } \{0, 1\} \quad
\text{D) } \{-2, 2\} \quad
\text{E) } \{-3, -1\}
\]
Solution:
Alternatively, square both sides to remove the absolute values safely ($|A|^2 = A^2$):
\[
(2x – 1)^2 = (x + 1)^2
\]
\[
4x^2 – 4x + 1 = x^2 + 2x + 1
\]
Move all terms to one side:
\[
3x^2 – 6x = 0 \Rightarrow 3x(x – 2) = 0
\]
\[
\Rightarrow x_1 = 0 \quad \text{and} \quad x_2 = 2
\]
\[
S = \{0, 2\}
\]
\(\textbf{Correct Answer: A} \)
QUESTION 21
What is the sum of the distinct values of $x$ satisfying the equation ?
\[ 2|x^2 – 2|^2 – |3x^2 – 6| – 2 = 0 \]
\[
\text{A) } -2 \quad
\text{B) } -1 \quad
\text{C) } 0 \quad
\text{D) } 1 \quad
\text{E) } 2
\]
Solution:
Factor out a 3 from the second term: $|3x^2 – 6| = |3(x^2 – 2)| = 3|x^2 – 2|$.
The equation stands as:
\[
2|x^2 – 2|^2 – 3|x^2 – 2| – 2 = 0
\]
Let $|x^2 – 2| = t$ (where $t \geq 0$):
\[
2t^2 – 3t – 2 = 0 \Rightarrow (2t + 1)(t – 2) = 0 \Rightarrow t = -\frac{1}{2} \quad \text{or} \quad t = 2
\]
Since $t$ represents an absolute value, it cannot be negative; discard $t = -\frac{1}{2}$.
Solve for $t = 2$:
\[
|x^2 – 2| = 2
\]
This yields two sub-cases:
1. $x^2 – 2 = 2 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$
2. $x^2 – 2 = -2 \Rightarrow x^2 = 0 \Rightarrow x = 0$
The set of distinct values is $\{-2, 0, 2\}$. Their sum is:
\[
-2 + 0 + 2 = 0
\]
\(\textbf{Correct Answer: C} \)
QUESTION 22
How many distinct real values of $x$ satisfy the equation ?
\[ |x^5 – 5x| = 11x \]
\[
\text{A) } 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]
Solution:
Because the left side $|x^5 – 5x| \geq 0$, the right side must also satisfy $11x \geq 0 \Rightarrow x \geq 0$. We split the absolute value equation:
Branch 1:
\[
x^5 – 5x = 11x \Rightarrow x^5 – 16x = 0
\]
\[
x(x^4 – 16) = 0 \Rightarrow x(x^2 – 4)(x^2 + 4) = 0
\]
\[
x(x – 2)(x + 2)(x^2 + 4) = 0 \Rightarrow x = 0, \quad x = 2, \quad x = -2
\]
Branch 2:
\[
x^5 – 5x = -11x \Rightarrow x^5 + 6x = 0
\]
\[
x(x^4 + 6) = 0 \Rightarrow x = 0 \quad (\text{since } x^4 + 6 \neq 0 \text{ for real numbers})
\]
Now evaluate these roots against our constraint $x \geq 0$:
– $x = 0$ (Valid)
– $x = 2$ (Valid)
– $x = -2$ (Invalid, since it is negative and would make $11x$ negative)
There are exactly $2$ distinct real solutions: $\{0, 2\}$.
\(\textbf{Correct Answer: B} \)
← Previous Page | Next Page →