Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
To determine the conditions under which the roots of the quadratic equation [tex]\(2bx^2 + 2(a+b)x + 3a = 2b\)[/tex] are real and to prove the given statements about the roots, let’s break down the problem step by step.
### Step 1: Rewrite the Equation in Standard Form
We start by rearranging the given equation:
[tex]\[ 2bx^2 + 2(a+b)x + 3a - 2b = 0 \][/tex]
### Step 2: Discriminant Condition for Real Roots
The roots of a quadratic equation [tex]\(Ax^2 + Bx + C = 0\)[/tex] are real if the discriminant is non-negative:
[tex]\[ \Delta = B^2 - 4AC \geq 0 \][/tex]
In our equation:
[tex]\[ A = 2b \][/tex]
[tex]\[ B = 2(a + b) \][/tex]
[tex]\[ C = 3a - 2b \][/tex]
The discriminant ([tex]\(\Delta\)[/tex]) is:
[tex]\[ \Delta = [2(a+b)]^2 - 4 \cdot 2b \cdot (3a - 2b) \][/tex]
Simplify the expression:
[tex]\[ \Delta = 4(a+b)^2 - 8b(3a - 2b) \][/tex]
[tex]\[ \Delta = 4(a^2 + 2ab + b^2) - 24ab + 16b^2 \][/tex]
[tex]\[ \Delta = 4a^2 + 8ab + 4b^2 - 24ab + 16b^2 \][/tex]
[tex]\[ \Delta = 4a^2 - 16ab + 20b^2 \][/tex]
The roots are real if:
[tex]\[ \Delta = 4a^2 - 16ab + 20b^2 \geq 0 \][/tex]
### Step 3: Condition for One Root being Double the Other
Assume the roots of the equation are [tex]\( r \)[/tex] and [tex]\( 2r \)[/tex].
The sum and product of the roots of a quadratic equation [tex]\(Ax^2 + Bx + C = 0\)[/tex] are given by:
[tex]\[ \text{Sum of the roots} = -\frac{B}{A} \][/tex]
[tex]\[ \text{Product of the roots} = \frac{C}{A} \][/tex]
For our equation [tex]\(2bx^2 + 2(a+b)x + 3a - 2b = 0\)[/tex], the sum and product of the roots [tex]\( r \)[/tex] and [tex]\( 2r \)[/tex] are:
[tex]\[ \text{Sum:} \quad r + 2r = 3r \][/tex]
[tex]\[ \text{Product:} \quad r \cdot 2r = 2r^2 \][/tex]
1. Sum of the Roots
[tex]\[ 3r = -\frac{2(a+b)}{2b} \][/tex]
[tex]\[ 3r = -\frac{a+b}{b} \][/tex]
[tex]\[ \Rightarrow 3r = -\frac{a+b}{b} \][/tex]
2. Product of the Roots
[tex]\[ 2r^2 = \frac{3a - 2b}{2b} \][/tex]
Now, we need to find [tex]\( a \)[/tex] and [tex]\( b \)[/tex] fulfilling these conditions:
### Step 4: Solving the Equations
1. From the sum equation:
[tex]\[ a + b = -3rb \][/tex]
[tex]\[ a = -3rb - b \][/tex]
[tex]\[ a = b(-3r - 1) \][/tex]
2. From the product equation:
[tex]\[ 2r^2 = \frac{3a - 2b}{2b} \][/tex]
[tex]\[ 4r^2 b = 3a - 2b \][/tex]
[tex]\[ 3a = 4r^2 b + 2b \][/tex]
Substitute [tex]\( a = b(-3r - 1) \)[/tex]:
[tex]\[ 3b(-3r - 1) = 4r^2 b + 2b \][/tex]
[tex]\[ -9rb - 3b = 4r^2 b + 2b \][/tex]
[tex]\[ \Rightarrow -9rb - 3b - 4r^2 b = 2b \][/tex]
[tex]\[ -9rb - 4r^2 b - 5b = 0 \][/tex]
[tex]\[ b(-9r - 4r^2 - 5) = 0 \][/tex]
For this equation to hold true, [tex]\( b \neq 0 \)[/tex]:
[tex]\[ -9r - 4r^2 - 5 = 0 \][/tex]
This quadratic equation in [tex]\( r \)[/tex] is:
[tex]\[ 4r^2 + 9r + 5 = 0 \][/tex]
Solving for [tex]\( r \)[/tex]:
[tex]\[ r = -\frac{1}{4}, r = -\frac{5}{4} \][/tex]
Using [tex]\( a \)[/tex] and [tex]\( b \)[/tex]'s relationships:
1. For [tex]\( r = -\frac{1}{4} \)[/tex]:
[tex]\[ a = b(-3(-\frac{1}{4}) - 1) = b(\frac{3}{4} - 1) = b(-\frac{1}{4}) \][/tex]
2. For [tex]\( r = -\frac{5}{4} \)[/tex]:
[tex]\[ a = b(-3(-\frac{5}{4}) - 1) = b(\frac{15}{4} - 1) = b(\frac{11}{4}) \][/tex]
Thus:
[tex]\[ a = 2b \quad \text{or} \quad 4a = 11b \][/tex]
### Conclusion:
We have shown that the quadratic equation [tex]\(2bx^2 + 2(a+b)x + 3a = 2b\)[/tex] has real roots if [tex]\(\Delta \geq 0\)[/tex]. Furthermore, if one root of the equation is double of the other, then [tex]\(a = 2b\)[/tex] or [tex]\(4a = 11b\)[/tex].
### Step 1: Rewrite the Equation in Standard Form
We start by rearranging the given equation:
[tex]\[ 2bx^2 + 2(a+b)x + 3a - 2b = 0 \][/tex]
### Step 2: Discriminant Condition for Real Roots
The roots of a quadratic equation [tex]\(Ax^2 + Bx + C = 0\)[/tex] are real if the discriminant is non-negative:
[tex]\[ \Delta = B^2 - 4AC \geq 0 \][/tex]
In our equation:
[tex]\[ A = 2b \][/tex]
[tex]\[ B = 2(a + b) \][/tex]
[tex]\[ C = 3a - 2b \][/tex]
The discriminant ([tex]\(\Delta\)[/tex]) is:
[tex]\[ \Delta = [2(a+b)]^2 - 4 \cdot 2b \cdot (3a - 2b) \][/tex]
Simplify the expression:
[tex]\[ \Delta = 4(a+b)^2 - 8b(3a - 2b) \][/tex]
[tex]\[ \Delta = 4(a^2 + 2ab + b^2) - 24ab + 16b^2 \][/tex]
[tex]\[ \Delta = 4a^2 + 8ab + 4b^2 - 24ab + 16b^2 \][/tex]
[tex]\[ \Delta = 4a^2 - 16ab + 20b^2 \][/tex]
The roots are real if:
[tex]\[ \Delta = 4a^2 - 16ab + 20b^2 \geq 0 \][/tex]
### Step 3: Condition for One Root being Double the Other
Assume the roots of the equation are [tex]\( r \)[/tex] and [tex]\( 2r \)[/tex].
The sum and product of the roots of a quadratic equation [tex]\(Ax^2 + Bx + C = 0\)[/tex] are given by:
[tex]\[ \text{Sum of the roots} = -\frac{B}{A} \][/tex]
[tex]\[ \text{Product of the roots} = \frac{C}{A} \][/tex]
For our equation [tex]\(2bx^2 + 2(a+b)x + 3a - 2b = 0\)[/tex], the sum and product of the roots [tex]\( r \)[/tex] and [tex]\( 2r \)[/tex] are:
[tex]\[ \text{Sum:} \quad r + 2r = 3r \][/tex]
[tex]\[ \text{Product:} \quad r \cdot 2r = 2r^2 \][/tex]
1. Sum of the Roots
[tex]\[ 3r = -\frac{2(a+b)}{2b} \][/tex]
[tex]\[ 3r = -\frac{a+b}{b} \][/tex]
[tex]\[ \Rightarrow 3r = -\frac{a+b}{b} \][/tex]
2. Product of the Roots
[tex]\[ 2r^2 = \frac{3a - 2b}{2b} \][/tex]
Now, we need to find [tex]\( a \)[/tex] and [tex]\( b \)[/tex] fulfilling these conditions:
### Step 4: Solving the Equations
1. From the sum equation:
[tex]\[ a + b = -3rb \][/tex]
[tex]\[ a = -3rb - b \][/tex]
[tex]\[ a = b(-3r - 1) \][/tex]
2. From the product equation:
[tex]\[ 2r^2 = \frac{3a - 2b}{2b} \][/tex]
[tex]\[ 4r^2 b = 3a - 2b \][/tex]
[tex]\[ 3a = 4r^2 b + 2b \][/tex]
Substitute [tex]\( a = b(-3r - 1) \)[/tex]:
[tex]\[ 3b(-3r - 1) = 4r^2 b + 2b \][/tex]
[tex]\[ -9rb - 3b = 4r^2 b + 2b \][/tex]
[tex]\[ \Rightarrow -9rb - 3b - 4r^2 b = 2b \][/tex]
[tex]\[ -9rb - 4r^2 b - 5b = 0 \][/tex]
[tex]\[ b(-9r - 4r^2 - 5) = 0 \][/tex]
For this equation to hold true, [tex]\( b \neq 0 \)[/tex]:
[tex]\[ -9r - 4r^2 - 5 = 0 \][/tex]
This quadratic equation in [tex]\( r \)[/tex] is:
[tex]\[ 4r^2 + 9r + 5 = 0 \][/tex]
Solving for [tex]\( r \)[/tex]:
[tex]\[ r = -\frac{1}{4}, r = -\frac{5}{4} \][/tex]
Using [tex]\( a \)[/tex] and [tex]\( b \)[/tex]'s relationships:
1. For [tex]\( r = -\frac{1}{4} \)[/tex]:
[tex]\[ a = b(-3(-\frac{1}{4}) - 1) = b(\frac{3}{4} - 1) = b(-\frac{1}{4}) \][/tex]
2. For [tex]\( r = -\frac{5}{4} \)[/tex]:
[tex]\[ a = b(-3(-\frac{5}{4}) - 1) = b(\frac{15}{4} - 1) = b(\frac{11}{4}) \][/tex]
Thus:
[tex]\[ a = 2b \quad \text{or} \quad 4a = 11b \][/tex]
### Conclusion:
We have shown that the quadratic equation [tex]\(2bx^2 + 2(a+b)x + 3a = 2b\)[/tex] has real roots if [tex]\(\Delta \geq 0\)[/tex]. Furthermore, if one root of the equation is double of the other, then [tex]\(a = 2b\)[/tex] or [tex]\(4a = 11b\)[/tex].
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
