Quadratic Equations
An equation of the form:
\[
ax^2 + bx + c = 0
\]
where $a, b, c \in \mathbb{R}$ and $a \neq 0$ is called a quadratic equation in one variable. The real numbers $x$ that satisfy this equation are called the roots of the equation, and the set formed by these roots is called the solution set.
The numbers $a, b, c \in \mathbb{R}$ are referred to as the coefficients of the quadratic equation.
Special Cases of the Equation \( ax^2 + bx + c = 0 \):
1) If \( b = 0 \) and \( c = 0 \):
\[
ax^2 = 0 \Rightarrow x^2 = 0
\]
\[
\Rightarrow x \cdot x = 0
\]
\[
\Rightarrow x = 0 \quad \text{or} \quad x = 0
\]
\[
\Rightarrow x_1 = x_2 = 0
\]
Thus, the equation has two equal roots (a double root).
The solution set is:
\[
S = \{ 0 \}
\]
2) If \( c = 0 \) and \( b \neq 0 \):
\[
ax^2 + bx = 0 \Rightarrow x (ax + b) = 0
\]
\[
\Rightarrow x = 0 \quad \text{or} \quad ax + b = 0
\]
\[
\Rightarrow x_1 = 0 \quad \text{or} \quad x_2 = -\frac{b}{a}
\]
The solution set is:
\[
S = \left\{ 0, -\frac{b}{a} \right\}
\]
Example:
Find the solution set of the equation $5x^2 – 3x = 0$.
\[
5x^2 – 3x = 0 \Rightarrow x (5x – 3) = 0
\]
\[
\Rightarrow x = 0 \quad \text{or} \quad 5x – 3 = 0
\]
\[
\Rightarrow x_1 = 0 \quad \text{or} \quad x_2 = \frac{3}{5}
\]
The solution set is found as:
\[
S = \left\{ 0, \frac{3}{5} \right\}
\]
3) If \( b = 0 \) and \( c \neq 0 \):
\[
ax^2 + c = 0 \Rightarrow x^2 = -\frac{c}{a}
\]
\( \bullet \quad \) If $a$ and $c$ have the **same sign**, then $-\frac{c}{a} < 0$. Since the square of a real number cannot be negative, the equation has no real roots. In this case, the solution set over the real numbers is: \[ S = \emptyset \] \( \bullet \quad \) If $a$ and $c$ have **opposite signs**, then $-\frac{c}{a} > 0$. The real roots of the equation are:
\[
x_1 = \sqrt{-\frac{c}{a}}, \quad x_2 = -\sqrt{-\frac{c}{a}}
\]
These roots have the same absolute value but opposite signs. Here, the sum of the roots is $x_1 + x_2 = 0$, and the solution set is:
\[
S = \left\{ \sqrt{-\frac{c}{a}}, -\sqrt{-\frac{c}{a}} \right\}
\]
Example:
Find the solution set of the equation $x^2 + 3 = 0$.
\[
x^2 + 3 = 0 \Rightarrow x^2 = -3
\]
Since there is no real number whose square is negative:
\[
S = \emptyset
\]
Example:
Find the solution set of the equation $3x^2 – 27 = 0$.
\[
3x^2 – 27 = 0 \Rightarrow x^2 = 9
\]
\[
\Rightarrow x_1 = -3 \quad \text{or} \quad x_2 = 3
\]
\[
\Rightarrow S = \{-3, 3\}
\]
Solving Equations of the Form \( ax^2 + bx + c = 0 \):
1) When the trinomial \( ax^2 + bx + c \) can be easily factored:
If it factors into the form:
\[
(px + m)(qx + n)
\]
then,
\[
ax^2 + bx + c = 0 \Rightarrow (px + m)(qx + n) = 0
\]
\[
\Rightarrow px + m = 0 \quad \text{or} \quad qx + n = 0
\]
\[ x_1 = -\frac{m}{p} \quad \text{and} \quad x_2 = -\frac{n}{q} \]
\[ S = \left\{ -\frac{m}{p}, \; -\frac{n}{q} \right\} \]
Example:
Find the solution set of the equation $x^2 – 998x – 2000 = 0$.
\[
x^2 – 998x – 2000 = 0 \Rightarrow (x + 2)(x – 1000) = 0
\]
\[
\Rightarrow x + 2 = 0 \quad \text{or} \quad x – 1000 = 0
\]
\[
\Rightarrow x_1 = -2 \quad \text{or} \quad x_2 = 1000
\]
\[
\Rightarrow S = \{-2, 1000\}
\]
Example:
Find the solution set of the equation $5x^2 + 99x – 20 = 0$.
\[
5x^2 + 99x – 20 = 0 \Rightarrow (5x – 1)(x + 20) = 0
\]
\[
\Rightarrow 5x – 1 = 0 \quad \text{or} \quad x + 20 = 0
\]
\[
\Rightarrow x_1 = \frac{1}{5} \quad \text{or} \quad x_2 = -20
\]
\[
\Rightarrow S = \left\{ -20, \frac{1}{5} \right\}
\]
Example:
Letting $a$ and $b$ be real numbers, find the solution set of the quadratic equation:
\[
ax^2 – (a^2b + b)x + ab^2 = 0
\]
\[
\Rightarrow (ax – b)(x – ab) = 0
\]
\[
\Rightarrow ax – b = 0 \quad \text{or} \quad x – ab = 0
\]
\[
\Rightarrow x_1 = \frac{b}{a} \quad \text{or} \quad x_2 = ab
\]
\[
\Rightarrow S = \left\{ \frac{b}{a}, \; ab \right\}
\]
Example:
Find the solution set of the equation $3x^2 + \sqrt{3}x – 2 = 0$.
\[
3x^2 + \sqrt{3}x – 2 = 0 \Rightarrow (\sqrt{3}x – 1)(\sqrt{3}x + 2) = 0
\]
\[
\Rightarrow \sqrt{3}x – 1 = 0 \quad \text{or} \quad \sqrt{3}x + 2 = 0
\]
\[
\Rightarrow x_1 = \frac{1}{\sqrt{3}} \quad \text{or} \quad x_2 = \frac{-2}{\sqrt{3}}
\]
\[
\Rightarrow S = \left\{ \frac{-2}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\}
\]
2) When the trinomial \( ax^2 + bx + c \) cannot be easily factored, we check the value \( b^2 – 4ac \).
This value is called the discriminant of the equation and is denoted by \( \Delta \) (Delta).
\( \bullet \quad \Delta = b^2 – 4ac > 0 \): The equation \( ax^2 + bx + c = 0 \) has two distinct real roots.
They are given by:
\[
x_1 = \frac{-b + \sqrt{\Delta}}{2a}, \quad x_2 = \frac{-b – \sqrt{\Delta}}{2a}
\]
Example:
Find the solution set of the equation $2x^2 + 3x – 1 = 0$.
\[
\Delta = b^2 – 4ac
\]
\[
= 3^2 – 4 \cdot 2 \cdot (-1) = 17 > 0
\]
Since $\Delta > 0$, the roots of the equation are:
\[
x_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-3 + \sqrt{17}}{2 \cdot 2} = \frac{-3 + \sqrt{17}}{4}
\]
\[
x_2 = \frac{-b – \sqrt{\Delta}}{2a} = \frac{-3 – \sqrt{17}}{2 \cdot 2} = \frac{-3 – \sqrt{17}}{4}
\]
The solution set is:
\[
S = \left\{ \frac{-3 + \sqrt{17}}{4}, \frac{-3 – \sqrt{17}}{4} \right\}
\]
Example:
Find the solution set of the equation $x^2 – \sqrt{5}x + 1 = 0$.
\[
\Delta = b^2 – 4ac
\]
\[
= (-\sqrt{5})^2 – 4 \cdot 1 \cdot 1 = 1 > 0
\]
Since $\Delta > 0$, the roots of the equation are:
\[
x_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-(-\sqrt{5}) + \sqrt{1}}{2 \cdot 1} = \frac{\sqrt{5} + 1}{2}
\]
\[
x_2 = \frac{-b – \sqrt{\Delta}}{2a} = \frac{-(-\sqrt{5}) – \sqrt{1}}{2 \cdot 1} = \frac{\sqrt{5} – 1}{2}
\]
The solution set is:
\[
S = \left\{ \frac{\sqrt{5} + 1}{2}, \frac{\sqrt{5} – 1}{2} \right\}
\]
Note:
If the coefficient $b$ in the equation \( ax^2 + bx + c = 0 \) is an even number, we can simplify calculations. Dividing both the numerator and the denominator of the quadratic formula roots by $2$, let:
\[
b’ = \frac{b}{2}
\]
Then we can define a modified discriminant:
\[
\Delta’ = (b’)^2 – ac
\]
In this case, the roots are calculated more simply as:
\[
x_{1,2} = \frac{-b’ \pm \sqrt{\Delta’}}{a}
\]
Example:
Find the solution set of the equation $x^2 + 32x + 56 = 0$.
\[
b’ = \frac{b}{2} = \frac{32}{2} = 16
\]
\[
\Delta’ = (b’)^2 – ac
\]
\[
= (16)^2 – 1 \cdot 56 = 200 > 0
\]
Since $\Delta’ > 0$, the roots are:
\[
x_1 = \frac{-b’ + \sqrt{\Delta’}}{a} = \frac{-16 + \sqrt{200}}{1} = -16 + 10\sqrt{2}
\]
\[
x_2 = \frac{-b’ – \sqrt{\Delta’}}{a} = \frac{-16 – \sqrt{200}}{1} = -16 – 10\sqrt{2}
\]
The solution set is:
\[
S = \left\{ -16 + 10\sqrt{2}, -16 – 10\sqrt{2} \right\}
\]
Note:
In any equation \( ax^2 + bx + c = 0 \), if \( a \) and \( c \) have opposite signs, then regardless of the value of \( b \), it is guaranteed that:
\[
\Delta = b^2 – 4ac > 0
\]
Examples:
\( \bullet \quad -3x^2 + x + 20 = 0 \)
\( \bullet \quad x^2 – 5x – 1000 = 0 \)
\( \bullet \quad -x^2 + 100x + 998 = 0 \)
For all the equations above, we can state that $\Delta > 0$ instantly without explicitly computing $b^2 – 4ac$.
b) If \( \Delta = b^2 – 4ac = 0 \):
The equation \( ax^2 + bx + c = 0 \) has two equal real roots (a double root or coincident roots).
They are calculated as:
\[
x_1 = x_2 = \frac{-b}{2a}
\]
Example:
Find the solution set of the equation $100x^2 – 20x + 1 = 0$.
\[
\Delta = b^2 – 4ac
\]
\[
= (-20)^2 – 4 \cdot 100 \cdot 1 = 0
\]
Since $\Delta = 0$, the equal roots are:
\[
x_1 = x_2 = \frac{-b}{2a} = \frac{-(-20)}{2 \cdot 100} = \frac{1}{10}
\]
The solution set is:
\[
S = \left\{ \frac{1}{10} \right\}
\]
Example:
Find the solution set of the equation $3x^2 – 4\sqrt{3}x + 4 = 0$.
\[
\Delta = b^2 – 4ac
\]
\[
= (-4\sqrt{3})^2 – 4 \cdot 3 \cdot 4 = 0
\]
Since $\Delta = 0$, the roots are:
\[
x_1 = x_2 = \frac{-b}{2a} = \frac{-(-4\sqrt{3})}{2 \cdot 3} = \frac{2\sqrt{3}}{3}
\]
The solution set is:
\[
S = \left\{ \frac{2\sqrt{3}}{3} \right\}
\]
Example:
Find the values of $m$ for which the equation $4mx^2 – (3m + 1)x + 1 = 0$ has two equal real roots.
For the equation to have two equal real roots, we must have:
\[
\Delta = b^2 – 4ac = 0
\]
\[
b^2 – 4ac = 0 \Rightarrow (-(3m+1))^2 – 4 \cdot 4m \cdot 1 = 0
\]
\[
\Rightarrow 9m^2 – 10m + 1 = 0
\]
\[
\Rightarrow (9m – 1)(m – 1) = 0
\]
\[
\Rightarrow m_1 = \frac{1}{9} \quad \text{or} \quad m_2 = 1
\]
QUESTION 1
Given that the equation $mx^2 + 2mx – x + m + 1 = 0$ has two equal real roots, what is the value of these roots?
\[
\text{A) } 2 \quad
\text{B) } 3 \quad
\text{C) } 4 \quad
\text{D) } 5 \quad
\text{E) } 6
\]
Solution:
Since the equation has equal roots, we require:
\[
\Delta = b^2 – 4ac = 0
\]
First, rearrange the given equation by grouping the $x$ terms:
\[
mx^2 + (2m – 1)x + m + 1 = 0
\]
Now evaluate the discriminant:
\[
\Delta = b^2 – 4ac = 0 \Rightarrow (2m – 1)^2 – 4m(m + 1) = 0
\]
\[
\Rightarrow 4m^2 – 4m + 1 – 4m^2 – 4m = 0 \Rightarrow -8m + 1 = 0 \Rightarrow m = \frac{1}{8}
\]
The equal roots are given by:
\[
x_1 = x_2 = \frac{-b}{2a} \Rightarrow x_1 = x_2 = \frac{-(2m – 1)}{2m}
\]
Substituting $m = \frac{1}{8}$:
\[
\Rightarrow x_1 = x_2 = \frac{-(2 \cdot \frac{1}{8} – 1)}{2 \cdot \frac{1}{8}} = \frac{-(\frac{1}{4} – 1)}{\frac{1}{4}} = \frac{\frac{3}{4}}{\frac{1}{4}} = 3
\]
\(\textbf{Correct Answer: B} \)
QUESTION 2
What must the value of $m$ be for the equation $3x^2 + (5m + 1)x + 2m^2 + m = 0$ to have a double root?
\[
\text{A) } 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]
Solution:
For the equation to have a double root:
\[
\Delta = b^2 – 4ac = 0 \Rightarrow (5m + 1)^2 – 4 \cdot 3 \cdot (2m^2 + m) = 0
\]
\[
\Rightarrow 25m^2 + 10m + 1 – 24m^2 – 12m = 0
\]
\[
\Rightarrow m^2 – 2m + 1 = 0
\]
\[
\Rightarrow (m – 1)^2 = 0 \Rightarrow m_1 = m_2 = 1
\]
\(\textbf{Correct Answer: A} \)
c) If \( \Delta = b^2 – 4ac < 0 \):
The equation \( ax^2 + bx + c = 0 \) has no real roots. Its solution set over the real numbers is empty:
\[
S = \emptyset
\]
Example:
Find the solution set of the equation $5x^2 + 7x + 3 = 0$.
\[
\Delta = b^2 – 4ac = 7^2 – 4 \cdot 5 \cdot 3 = 49 – 60 = -11 < 0
\]
Since the discriminant is negative, the solution set over the real numbers is:
\[
S = \emptyset
\]
QUESTION 3
In which interval must $m$ lie for the equation $mx^2 + 2\sqrt{2}x – 1 = 0$ to have no real roots?
\[
\text{A) } (3, 4) \quad
\text{B) } (-2, 0) \quad
\text{C) } (-1, 1) \quad
\text{D) } (-\infty, -2) \quad
\text{E) } (1, \infty)
\]
Solution:
For the equation to have no real roots, we must satisfy $\Delta < 0$:
\[
\Delta = b^2 – 4ac < 0
\]
\[
\Rightarrow (2\sqrt{2})^2 – 4 \cdot m \cdot (-1) < 0
\]
\[
\Rightarrow 8 + 4m < 0
\]
\[
\Rightarrow 4m < -8 \Rightarrow m < -2
\]
Hence, $m \in (-\infty, -2)$.
\(\textbf{Correct Answer: D} \)
Note:
If the sum of the coefficients of an $n$-th degree single-variable polynomial equation is $0$ (zero), then $1$ is a root of this equation.
Therefore, if the sum of the coefficients of the quadratic equation \( ax^2 + bx + c = 0 \) is zero (\( a + b + c = 0 \)), then one root is explicitly \( x_1 = 1 \), and the other root is \( x_2 = \frac{c}{a} \).
Accordingly, the solution set for such equations can be written directly as:
\[
S = \left\{ 1, \frac{c}{a} \right\}
\]
Example:
Given that one of the roots of the equation $x^2 – (5m + 4)x + m^2 – m – 1 = 0$ is $x_1 = -1$, find the other root.
Since $-1$ is a root, it must satisfy the equation. Substituting $x = -1$:
\[
(-1)^2 – (5m + 4)(-1) + m^2 – m – 1 = 0
\]
\[
\Rightarrow 1 + 5m + 4 + m^2 – m – 1 = 0
\]
\[
\Rightarrow m^2 + 4m + 4 = 0
\]
\[
\Rightarrow (m + 2)^2 = 0 \Rightarrow m = -2
\]
Now, substitute $m = -2$ back into the original equation:
\[
x^2 – (5 \cdot (-2) + 4)x + (-2)^2 – (-2) – 1 = 0
\]
\[
\Rightarrow x^2 – (-6)x + 4 + 2 – 1 = 0
\]
\[
\Rightarrow x^2 + 6x + 5 = 0
\]
Factoring the equation:
\[
\Rightarrow (x + 1)(x + 5) = 0
\]
\[
\Rightarrow x_1 = -1, \quad x_2 = -5
\]
The other root is $-5$.
Example:
Let $m$ be a real number. We can immediately deduce that one root of the equation:
\[
(3m + 2)x^2 – (m^2 + m)x + m^2 – 2m – 2 = 0
\]
is \( x_1 = 1 \) because the sum of its coefficients is:
\[
(3m + 2) + [-(m^2 + m)] + (m^2 – 2m – 2) = 3m + 2 – m^2 – m + m^2 – 2m – 2 = 0
\]
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