Answer :
Let's start with question 13.
### 13. If [tex]\((k+2)x^2 - (k+2)x - 1 = 0\)[/tex] has one real root, then the value of [tex]\(k\)[/tex] is:
To determine the values of [tex]\(k\)[/tex] for which the given quadratic equation has one real root, we need to set the discriminant of the quadratic equation to zero. The general form of a quadratic equation is [tex]\(ax^2 + bx + c = 0\)[/tex], with the discriminant given by [tex]\(\Delta = b^2 - 4ac\)[/tex].
For the given equation [tex]\((k+2)x^2 - (k+2)x - 1 = 0\)[/tex], let's identify the coefficients:
- [tex]\(a = k + 2\)[/tex]
- [tex]\(b = -(k + 2)\)[/tex]
- [tex]\(c = -1\)[/tex]
The discriminant [tex]\(\Delta\)[/tex] is calculated as:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values:
[tex]\[ \Delta = [-(k+2)]^2 - 4(k+2)(-1) \][/tex]
Simplify the discriminant:
[tex]\[ \Delta = (k+2)^2 - 4(k+2)(-1) \][/tex]
[tex]\[ \Delta = (k+2)^2 + 4(k+2) \][/tex]
[tex]\[ \Delta = (k+2)^2 + 4(k+2) \][/tex]
[tex]\[ \Delta = (k+2)[(k+2) + 4] \][/tex]
[tex]\[ \Delta = (k+2)(k+2+4) \][/tex]
[tex]\[ \Delta = (k+2)(k+6) \][/tex]
For the quadratic equation to have exactly one real root, the discriminant must be equal to zero:
[tex]\[ (k+2)(k+6) = 0 \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ k+2 = 0 \quad \text{or} \quad k+6 = 0 \][/tex]
[tex]\[ k = -2 \quad \text{or} \quad k = -6 \][/tex]
Thus, the values of [tex]\(k\)[/tex] that make the equation have one real root are [tex]\(-2\)[/tex] and [tex]\(-6\)[/tex]. The answer is:
C. -6 or -2
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### 14. The simplified form of [tex]\(\frac{(a+b)^2 - 1}{a+b+1}\)[/tex] is:
First, let's rewrite the numerator and the denominator:
[tex]\[ \text{Numerator: } (a+b)^2 - 1 \][/tex]
[tex]\[ \text{Denominator: } a+b+1 \][/tex]
The numerator [tex]\((a+b)^2 - 1\)[/tex] can be factored as a difference of squares:
[tex]\[ (a+b)^2 - 1 = (a+b)^2 - 1^2 = [(a+b) - 1][(a+b) + 1] \][/tex]
[tex]\[ (a+b)^2 - 1 = (a+b-1)(a+b+1) \][/tex]
So, the expression becomes:
[tex]\[ \frac{(a+b-1)(a+b+1)}{a+b+1} \][/tex]
We can cancel out the common factor [tex]\(a+b+1\)[/tex] in the numerator and the denominator:
[tex]\[ \frac{(a+b-1)(a+b+1)}{a+b+1} = a+b-1 \][/tex]
Thus, the simplified form of the expression is:
D. [tex]\(a+b-1\)[/tex]
### 13. If [tex]\((k+2)x^2 - (k+2)x - 1 = 0\)[/tex] has one real root, then the value of [tex]\(k\)[/tex] is:
To determine the values of [tex]\(k\)[/tex] for which the given quadratic equation has one real root, we need to set the discriminant of the quadratic equation to zero. The general form of a quadratic equation is [tex]\(ax^2 + bx + c = 0\)[/tex], with the discriminant given by [tex]\(\Delta = b^2 - 4ac\)[/tex].
For the given equation [tex]\((k+2)x^2 - (k+2)x - 1 = 0\)[/tex], let's identify the coefficients:
- [tex]\(a = k + 2\)[/tex]
- [tex]\(b = -(k + 2)\)[/tex]
- [tex]\(c = -1\)[/tex]
The discriminant [tex]\(\Delta\)[/tex] is calculated as:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values:
[tex]\[ \Delta = [-(k+2)]^2 - 4(k+2)(-1) \][/tex]
Simplify the discriminant:
[tex]\[ \Delta = (k+2)^2 - 4(k+2)(-1) \][/tex]
[tex]\[ \Delta = (k+2)^2 + 4(k+2) \][/tex]
[tex]\[ \Delta = (k+2)^2 + 4(k+2) \][/tex]
[tex]\[ \Delta = (k+2)[(k+2) + 4] \][/tex]
[tex]\[ \Delta = (k+2)(k+2+4) \][/tex]
[tex]\[ \Delta = (k+2)(k+6) \][/tex]
For the quadratic equation to have exactly one real root, the discriminant must be equal to zero:
[tex]\[ (k+2)(k+6) = 0 \][/tex]
Solving for [tex]\(k\)[/tex]:
[tex]\[ k+2 = 0 \quad \text{or} \quad k+6 = 0 \][/tex]
[tex]\[ k = -2 \quad \text{or} \quad k = -6 \][/tex]
Thus, the values of [tex]\(k\)[/tex] that make the equation have one real root are [tex]\(-2\)[/tex] and [tex]\(-6\)[/tex]. The answer is:
C. -6 or -2
---
### 14. The simplified form of [tex]\(\frac{(a+b)^2 - 1}{a+b+1}\)[/tex] is:
First, let's rewrite the numerator and the denominator:
[tex]\[ \text{Numerator: } (a+b)^2 - 1 \][/tex]
[tex]\[ \text{Denominator: } a+b+1 \][/tex]
The numerator [tex]\((a+b)^2 - 1\)[/tex] can be factored as a difference of squares:
[tex]\[ (a+b)^2 - 1 = (a+b)^2 - 1^2 = [(a+b) - 1][(a+b) + 1] \][/tex]
[tex]\[ (a+b)^2 - 1 = (a+b-1)(a+b+1) \][/tex]
So, the expression becomes:
[tex]\[ \frac{(a+b-1)(a+b+1)}{a+b+1} \][/tex]
We can cancel out the common factor [tex]\(a+b+1\)[/tex] in the numerator and the denominator:
[tex]\[ \frac{(a+b-1)(a+b+1)}{a+b+1} = a+b-1 \][/tex]
Thus, the simplified form of the expression is:
D. [tex]\(a+b-1\)[/tex]
