Answer :
To determine the value of [tex]\( k \)[/tex] for which the quadratic equation [tex]\((k+4) x^2 + (k+1) x + 1 = 0\)[/tex] has equal roots, we need to examine the discriminant of the quadratic equation.
The general form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are coefficients.
For our given quadratic equation:
- [tex]\( a = k+4 \)[/tex]
- [tex]\( b = k+1 \)[/tex]
- [tex]\( c = 1 \)[/tex]
The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
For the roots to be equal, the discriminant must be zero:
[tex]\[ \Delta = 0 \][/tex]
Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the discriminant formula:
[tex]\[ \Delta = (k+1)^2 - 4(k+4)(1) \][/tex]
Set the discriminant equal to zero:
[tex]\[ (k+1)^2 - 4(k+4) = 0 \][/tex]
Now, let's solve this equation step-by-step.
1. Expand and simplify the equation:
[tex]\[ (k + 1)^2 = k^2 + 2k + 1 \][/tex]
[tex]\[ 4(k + 4) = 4k + 16 \][/tex]
2. Substitute these expressions back into the equation and set it to zero:
[tex]\[ k^2 + 2k + 1 - 4k - 16 = 0 \][/tex]
3. Simplify further:
[tex]\[ k^2 + 2k + 1 - 4k - 16 = k^2 - 2k - 15 = 0 \][/tex]
4. Now, solve the quadratic equation [tex]\( k^2 - 2k - 15 = 0 \)[/tex].
To factorize:
[tex]\[ k^2 - 2k - 15 = (k - 5)(k + 3) = 0 \][/tex]
Thus, the solutions to the equation are:
[tex]\[ k - 5 = 0 \quad \text{or} \quad k + 3 = 0 \][/tex]
[tex]\[ k = 5 \quad \text{or} \quad k = -3 \][/tex]
Therefore, the values of [tex]\( k \)[/tex] for which the quadratic equation [tex]\((k+4)x^2 + (k+1)x + 1 = 0\)[/tex] has equal roots are:
[tex]\[ \boxed{5 \text{ and } -3} \][/tex]
The general form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are coefficients.
For our given quadratic equation:
- [tex]\( a = k+4 \)[/tex]
- [tex]\( b = k+1 \)[/tex]
- [tex]\( c = 1 \)[/tex]
The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
For the roots to be equal, the discriminant must be zero:
[tex]\[ \Delta = 0 \][/tex]
Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the discriminant formula:
[tex]\[ \Delta = (k+1)^2 - 4(k+4)(1) \][/tex]
Set the discriminant equal to zero:
[tex]\[ (k+1)^2 - 4(k+4) = 0 \][/tex]
Now, let's solve this equation step-by-step.
1. Expand and simplify the equation:
[tex]\[ (k + 1)^2 = k^2 + 2k + 1 \][/tex]
[tex]\[ 4(k + 4) = 4k + 16 \][/tex]
2. Substitute these expressions back into the equation and set it to zero:
[tex]\[ k^2 + 2k + 1 - 4k - 16 = 0 \][/tex]
3. Simplify further:
[tex]\[ k^2 + 2k + 1 - 4k - 16 = k^2 - 2k - 15 = 0 \][/tex]
4. Now, solve the quadratic equation [tex]\( k^2 - 2k - 15 = 0 \)[/tex].
To factorize:
[tex]\[ k^2 - 2k - 15 = (k - 5)(k + 3) = 0 \][/tex]
Thus, the solutions to the equation are:
[tex]\[ k - 5 = 0 \quad \text{or} \quad k + 3 = 0 \][/tex]
[tex]\[ k = 5 \quad \text{or} \quad k = -3 \][/tex]
Therefore, the values of [tex]\( k \)[/tex] for which the quadratic equation [tex]\((k+4)x^2 + (k+1)x + 1 = 0\)[/tex] has equal roots are:
[tex]\[ \boxed{5 \text{ and } -3} \][/tex]