Relationships Between Roots and Coefficients

 

\[
ax^2 + bx + c = 0
\]

Given that the roots of the equation are

\[
x_1 = \frac{-b + \sqrt{\Delta}}{2a} \quad \text{and} \quad x_2 = \frac{-b – \sqrt{\Delta}}{2a}
\]

the following relations hold:

 

Sum of the Roots:

 

\[
x_1 + x_2 = \frac{-b}{a}
\]

 

Product of the Roots:

 

\[
x_1 \cdot x_2 = \frac{c}{a}
\]

 

Absolute Value of the Difference of the Roots:

 

\[
|x_1 – x_2| = \frac{\sqrt{\Delta}}{|a|}
\]

By utilizing these fundamental relations and algebraic identities from factoring, we can derive the expressions below.

 

Sum of the Reciprocals of the Roots:

 

\[
\frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2}{x_1 \cdot x_2}
\]

Since this identity holds, we obtain:

\[
\frac{1}{x_1} + \frac{1}{x_2} = \frac{-b}{c}
\]

 

Sum of the Squares of the Roots:

 

\[
x_1^2 + x_2^2 = (x_1 + x_2)^2 – 2x_1 \cdot x_2 \quad \text{Since}
\]

\[
x_1^2 + x_2^2 = \frac{b^2 – 2ac}{a^2}
\]

 

Sum of the Cubes of the Roots:

 

\[
x_1^3 + x_2^3 = (x_1 + x_2)^3 – 3x_1 x_2 \cdot (x_1 + x_2)
\]

Since this identity holds, we obtain:

\[
x_1^3 + x_2^3 = \frac{3abc – b^3}{a^3}
\]

 

 

Example:

 

\[
2x^2 + 6x + 3 = 0
\]

Let \( x_1, x_2 \) be the roots of the equation.

\( \bullet \quad x_1 + x_2 = \frac{-b}{a} = \frac{-6}{2} = -3 \)

\( \bullet \quad x_1 \cdot x_2 = \frac{c}{a} = \frac{3}{2} \)

\( \bullet \quad |x_1 – x_2| = \frac{\sqrt{\Delta}}{|a|} = \frac{\sqrt{12}}{|2|} = \frac{2\sqrt{3}}{2} = \sqrt{3} \)

 

Example:

 

\[ x^2 + x – 3 = 0 \] Let \( x_1, x_2 \) be the roots of the equation. Find the value of the sum \( x_1^6 + x_2^6 \).

 

\[
x_1^6 + x_2^6 = (x_1^3 + x_2^3)^2 – 2 (x_1 \cdot x_2)^3
\]

\[
= \left( \frac{3abc – b^3}{a^3} \right)^2 – 2 \left( \frac{c}{a} \right)^3
\]

\[
= \left( \frac{-9 – 1}{1} \right)^2 – 2 \cdot \left( \frac{-3}{1} \right)^3
\]

\[
= 154
\]

 

QUESTION 23

 

\[
-x^2 + 3x – 1 = 0
\]

The roots of the equation are \( x_1, x_2 \). If \( x_1 > x_2 \), what is the value of the expression?

\[
x_1^3 \cdot x_2 – x_1 \cdot x_2^3
\]

 

\[
\text{A) } \sqrt{ 5} \quad
\text{B) } 2\sqrt{ 5} \quad
\text{C) } 3\sqrt{ 5} \quad
\text{D) } 4\sqrt{ 5} \quad
\text{E) } 5\sqrt{ 5}
\]

 

Solution:

 

\[
x_1^3 \cdot x_2 – x_1 \cdot x_2^3 = x_1 \cdot x_2 \cdot (x_1 – x_2) \cdot (x_1 + x_2)
\]

\[
= \frac{c}{a} \cdot \frac{\sqrt{\Delta}}{|a|} \cdot \left(-\frac{b}{a}\right)
\]

\[
= 1 \cdot \sqrt{5} \cdot 3 = 3\sqrt{5}
\]

 

\(\textbf{Answer: C} \)

 

QUESTION 24

 

\[
mx^2 – 3mx + 1 = 0
\]

The roots of the equation are \( x_1, x_2 \).  If \( 3x_1 \ – \ x_2 = 5 \), what is the value of \( m \)?

 

\[
\text{A) } \frac{1}{2} \quad
\text{B) } 1 \quad
\text{C) } \frac{3}{2} \quad
\text{D) } 2 \quad
\text{E) } 3
\]

 

Solution:

 

\[
x_1 + x_2 = \frac{-b}{a} = \frac{-(-3m)}{m} = 3
\]

Solving this simultaneously with the given equation,

\[
3x_1 – x_2 = 5
\]

yields \( x_1 = 2 \). Substituting this root into the original equation:

\[
m x^2 – 3m x + 1 = 0 \Rightarrow m \cdot 2^2 – 3m \cdot 2 + 1 = 0
\]

\[
\Rightarrow m = \frac{1}{2}
\]

 

\(\textbf{Answer: A} \)

 

QUESTION 25

 

\[
x^2 + mx + 2 = 0
\]

The roots of the equation are \( x_1, x_2 \). If \( x_1 < x_2 \) and

\[
x_1 + \frac{x_1}{x_2} = 1
\]

find the positive value of \( m \).

\[
\text{A) } 2 \sqrt{3 } \quad
\text{B) } \sqrt{3 } \quad
\text{C) } 3 \quad
\text{D) } 2 \quad
\text{E) } 1
\]

 

Solution:

 

\[
x_1 + \frac{x_1}{x_2} = 1 \Rightarrow x_1 \cdot x_2 + x_1 = x_2
\]

\[
\Rightarrow 2 + x_1 = x_2
\]

\[
\Rightarrow x_2 – x_1 = 2
\]

\[ \Rightarrow \sqrt{m^2 \;-\;8} = 2 \]

\[ m_1=-2 \sqrt{3} \quad \text{and } \quad m_2= 2 \sqrt{3} \]

 

\(\textbf{Answer: A} \)

 

QUESTION 26

 

Let \( a \) and \( b \) be non-zero real numbers. Given that the roots of the equation

\[
x^2 \ – \  (3a – b)x + a + b = 0
\]

are \( a \) and \( b \), find the value of the expression \( a^2 + b^2 \).

\[
\text{A) } 4 \quad
\text{B) } 5 \quad
\text{C) } 6 \quad
\text{D) } 7 \quad
\text{E) } 8
\]

 

Solution:

 

Using the product of the roots:

\[
a \cdot b = a + b
\]

and the sum of the roots:

\[
a + b = 3a \ – \  b
\]

\[
\Rightarrow a = b
\]

Substituting this into the product equation gives:

\[
a \cdot b = a + b \Rightarrow a^2 = 2a
\]

\[
\Rightarrow a = 2 \quad \text{and} \quad b = 2
\]

Thus,

\[
a^2 + b^2 = 8
\]

 

\(\textbf{Answer: E} \)

 

QUESTION 27

 

\[
mx^2 \ – \  x + 2m + 1 = 0
\]

By how much does the product of the roots of the given equation exceed the sum of its roots?

\[
\text{A) } 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]

 

Solution:

 

Let \( x_1, x_2 \) be the roots of the equation.

\[
x_1 \cdot x_2 = \frac{2m + 1}{m} = 2 + \frac{1}{m}
\]

and

\[
x_1 + x_2 = \frac{1}{m}
\]

Subtracting the sum from the product yields:

\[
x_1 \cdot x_2 \ – \  (x_1 + x_2) = 2 + \frac{1}{m}\; -\; \frac{1}{m} = 2
\]

 

\(\textbf{Answer: B} \)

 

QUESTION 28

 

\[
x^2 \ – \  6x + 1 = 0
\]

The roots of the equation are \( x_1 \) and \( x_2 \). Find the value of the expression

\[
\sqrt{x_1} + \sqrt{x_2}
\]

.

\[
\text{A) } 1 \quad
\text{B) } 2 \quad
\text{C) } \sqrt{2 } \quad
\text{D) } 2\sqrt{2 } \quad
\text{E) } 3\sqrt{2 }
\]

 

Solution:

 

Let

\[
t = \sqrt{x_1} + \sqrt{x_2}
\]

Squaring both sides gives:

\[
t^2 = x_1 + x_2 + 2\sqrt{x_1 x_2} \Rightarrow t^2 = 6 + 2\sqrt{1}
\]

\[
\Rightarrow t^2 = 8
\]

\[
\Rightarrow t = 2\sqrt{2} \quad \text{or} \quad t = -2\sqrt{2}
\]

Since the principal square roots must be positive,

\[
t = \sqrt{x_1} + \sqrt{x_2} > 0
\]

Therefore,

\[
\sqrt{x_1} + \sqrt{x_2} = 2\sqrt{2}
\]

 

\(\textbf{Answer: D} \)

 

QUESTION 29

 

One root of the equation

\[
x^2 \  – \  mx + n = 0
\]

is exactly 3 times a root of the equation

\[
x^2 + nx + m = 0
\]

. Given that the remaining roots of both equations are equal, which of the following represents the ratio of the roots of the equation?

\[
x^2 + nx + m = 0
\]

 

\[
\text{A) } 1 \quad
\text{B) } -\frac{1}{2} \quad
\text{C) } \frac{1}{2} \quad
\text{D) } -\frac{2}{5} \quad
\text{E) } \frac{2}{5}
\]

 

Solution:

 

Let the roots of the equation \( x^2 + nx + m = 0 \) be \( a \) and \( b \). Then, the roots of the equation \( x^2 – mx + n = 0 \) must be \( 3a \) and \( b \). Using Vieta’s formulas for the product and sum of the roots:

\[
a \cdot b = m
\]

\[
3a \cdot b = n
\]

\[
\Rightarrow n = 3m
\]

For the sum of the roots:

\[
a + b = -{n} = -3m
\]

\[ 3a+b=m \]

Solving this system of equations for \( a \) and \( b \):

\[ a= 2m \quad \text{and } \quad b=-5m \]

\[
\Rightarrow a = 2m \quad \text{and} \quad b = -5m
\]

Thus, the ratio of the roots is:

\[
\frac{a}{b} = \frac{2m}{-5m} = -\frac{2}{5}
\]

\(\textbf{Answer: D} \)

 

QUESTION 30

 

The roots of the equation

\[
x^2 – (3m + 1)x + 7m + 3 = 0
\]

are twice the roots of the equation

\[
x^2 – (m + 2)x + 2m = 0
\]

. What is the product of the roots of the equation ?

\[
x^2 – (3m + 1)x + 7m + 3 = 0
\]

 

\[
\text{A) } 16 \quad
\text{B) } 18 \quad
\text{C) } 20 \quad
\text{D) } 22 \quad
\text{E) } 24
\]

 

Solution:

 

Let the roots of the equation \( x^2 – (m + 2)x + 2m = 0 \) be \( a \) and \( b \). Then, the roots of the equation \( x^2 – (3m + 1)x + 7m + 3 = 0 \) are \( 2a \) and \( 2b \). Using the sum of the roots:

\[
a + b = m + 2
\]

\[
2a + 2b = 3m + 1
\]

\[
\Rightarrow 2(m + 2) = 3m + 1
\]

\[
\Rightarrow m = 3
\]

Substituting \( m = 3 \) into the target equation:

\[
x^2 – 10x + 24 = 0
\]

The product of the roots of this equation is given by:

\[
24
\]

 

\(\textbf{Answer: E} \)

 

 

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