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To determine the value of [tex]\( k \)[/tex] such that the given equation is true for all real values of [tex]\( x \)[/tex]:
[tex]\[ \left(a^3-3a-2\right)x^2 + \left(a^2-a-2\right)x = 2a^4 - 10a + k, \][/tex]
we need to compare the coefficients of corresponding powers of [tex]\( x \)[/tex] from both sides of the equation. The equation must hold for all real [tex]\( x \)[/tex], so the coefficients for [tex]\( x^2 \)[/tex], [tex]\( x \)[/tex], and the constant term must match on both sides.
### Step 1: Coefficient of [tex]\( x^2 \)[/tex]
The coefficient of [tex]\( x^2 \)[/tex] on the left-hand side (LHS) is:
[tex]\[ a^3 - 3a - 2 \][/tex]
There is no [tex]\( x^2 \)[/tex] term on the right-hand side (RHS) which implies the coefficient must be zero:
[tex]\[ a^3 - 3a - 2 = 0 \][/tex]
### Step 2: Coefficient of [tex]\( x \)[/tex]
The coefficient of [tex]\( x \)[/tex] on the LHS is:
[tex]\[ a^2 - a - 2 \][/tex]
There is no [tex]\( x \)[/tex] term on the RHS which implies the coefficient must be zero:
[tex]\[ a^2 - a - 2 = 0 \][/tex]
### Step 3: Constant Term
Since there is no constant term (independent of [tex]\( x \)[/tex]) on the LHS, it must be zero. The RHS constant term is:
[tex]\[ 2a^4 - 10a + k = 0 \][/tex]
### Solving the equations
First, solve the equations from Step 1 and Step 2 to find possible values of [tex]\( a \)[/tex].
1. [tex]\( a^3 - 3a - 2 = 0 \)[/tex]
2. [tex]\( a^2 - a - 2 = 0 \)[/tex]
Solve the quadratic equation [tex]\( a^2 - a - 2 = 0 \)[/tex]:
[tex]\[ a^2 - a - 2 = (a - 2)(a + 1) = 0 \][/tex]
Thus, [tex]\( a = 2 \)[/tex] or [tex]\( a = -1 \)[/tex].
Check which of these values satisfy [tex]\( a^3 - 3a - 2 = 0 \)[/tex]:
For [tex]\( a = 2 \)[/tex]:
[tex]\[ 2^3 - 3(2) - 2 = 8 - 6 - 2 = 0 \quad \text{(True)} \][/tex]
For [tex]\( a = -1 \)[/tex]:
[tex]\[ (-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0 \quad \text{(True)} \][/tex]
Both [tex]\( a = 2 \)[/tex] and [tex]\( a = -1 \)[/tex] are solutions.
### Determining [tex]\( k \)[/tex]
#### For [tex]\( a = 2 \)[/tex]:
[tex]\[ 2a^4 - 10a + k = 2(2)^4 - 10(2) + k = 2(16) - 20 + k = 32 - 20 + k = 0 \][/tex]
[tex]\[ k = -12 \][/tex]
#### For [tex]\( a = -1 \)[/tex]:
[tex]\[ 2a^4 - 10a + k = 2(-1)^4 - 10(-1) + k = 2(1) + 10 + k = 2 + 10 + k = 0 \][/tex]
[tex]\[ k = -12 \][/tex]
Thus, the value of [tex]\( k \)[/tex] that satisfies the equation for all real [tex]\( x \)[/tex] is:
[tex]\[ \boxed{-12} \][/tex]
[tex]\[ \left(a^3-3a-2\right)x^2 + \left(a^2-a-2\right)x = 2a^4 - 10a + k, \][/tex]
we need to compare the coefficients of corresponding powers of [tex]\( x \)[/tex] from both sides of the equation. The equation must hold for all real [tex]\( x \)[/tex], so the coefficients for [tex]\( x^2 \)[/tex], [tex]\( x \)[/tex], and the constant term must match on both sides.
### Step 1: Coefficient of [tex]\( x^2 \)[/tex]
The coefficient of [tex]\( x^2 \)[/tex] on the left-hand side (LHS) is:
[tex]\[ a^3 - 3a - 2 \][/tex]
There is no [tex]\( x^2 \)[/tex] term on the right-hand side (RHS) which implies the coefficient must be zero:
[tex]\[ a^3 - 3a - 2 = 0 \][/tex]
### Step 2: Coefficient of [tex]\( x \)[/tex]
The coefficient of [tex]\( x \)[/tex] on the LHS is:
[tex]\[ a^2 - a - 2 \][/tex]
There is no [tex]\( x \)[/tex] term on the RHS which implies the coefficient must be zero:
[tex]\[ a^2 - a - 2 = 0 \][/tex]
### Step 3: Constant Term
Since there is no constant term (independent of [tex]\( x \)[/tex]) on the LHS, it must be zero. The RHS constant term is:
[tex]\[ 2a^4 - 10a + k = 0 \][/tex]
### Solving the equations
First, solve the equations from Step 1 and Step 2 to find possible values of [tex]\( a \)[/tex].
1. [tex]\( a^3 - 3a - 2 = 0 \)[/tex]
2. [tex]\( a^2 - a - 2 = 0 \)[/tex]
Solve the quadratic equation [tex]\( a^2 - a - 2 = 0 \)[/tex]:
[tex]\[ a^2 - a - 2 = (a - 2)(a + 1) = 0 \][/tex]
Thus, [tex]\( a = 2 \)[/tex] or [tex]\( a = -1 \)[/tex].
Check which of these values satisfy [tex]\( a^3 - 3a - 2 = 0 \)[/tex]:
For [tex]\( a = 2 \)[/tex]:
[tex]\[ 2^3 - 3(2) - 2 = 8 - 6 - 2 = 0 \quad \text{(True)} \][/tex]
For [tex]\( a = -1 \)[/tex]:
[tex]\[ (-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0 \quad \text{(True)} \][/tex]
Both [tex]\( a = 2 \)[/tex] and [tex]\( a = -1 \)[/tex] are solutions.
### Determining [tex]\( k \)[/tex]
#### For [tex]\( a = 2 \)[/tex]:
[tex]\[ 2a^4 - 10a + k = 2(2)^4 - 10(2) + k = 2(16) - 20 + k = 32 - 20 + k = 0 \][/tex]
[tex]\[ k = -12 \][/tex]
#### For [tex]\( a = -1 \)[/tex]:
[tex]\[ 2a^4 - 10a + k = 2(-1)^4 - 10(-1) + k = 2(1) + 10 + k = 2 + 10 + k = 0 \][/tex]
[tex]\[ k = -12 \][/tex]
Thus, the value of [tex]\( k \)[/tex] that satisfies the equation for all real [tex]\( x \)[/tex] is:
[tex]\[ \boxed{-12} \][/tex]
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