Answer :
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].
