Find the value of k for which the quadratic equation (k+1)x^2-6(k+1)x+3(k+9)=0,k≠-1 has equal roots

General
1

Find the value of k for which the quadratic equation (k+1)x^2-6(k+1)x+3(k+9)=0,k≠-1 has equal roots.​

2

Find the value of ( k ) for which the quadratic equation (k+1)x^2 - 6(k+1)x + 3(k+9) = 0, ( k \neq -1 ), has equal roots.

Answer:
To determine the value of ( k ) for which the given quadratic equation has equal roots, we need to use the concept of the discriminant. For a quadratic equation of the form ( ax^2 + bx + c = 0 ), the discriminant is given by:

\Delta = b^2 - 4ac

Equal roots occur if and only if the discriminant (( \Delta )) is zero.

Solution By Steps:

  1. Identify coefficients ( a ), ( b ), and ( c ):

    From the equation ( (k+1)x^2 - 6(k+1)x + 3(k+9) = 0 ), we have:

    • ( a = k+1 )
    • ( b = -6(k+1) )
    • ( c = 3(k+9) )

  2. Calculate the discriminant:

    Substitute ( a ), ( b ), and ( c ) into the discriminant formula:

    \Delta = b^2 - 4ac

    Given ( b = -6(k+1) ):

    b^2 = [-6(k+1)]^2 = 36(k+1)^2

    And since ( a = k+1 ) and ( c = 3(k+9) ):

    4ac = 4(k+1) \cdot 3(k+9) = 12(k+1)(k+9)

    Therefore, the discriminant is:

    \Delta = 36(k+1)^2 - 12(k+1)(k+9)

  3. Set the discriminant to zero and solve for ( k ):

    36(k+1)^2 - 12(k+1)(k+9) = 0

    Factor out the common term ( 12(k+1) ):

    12(k+1)[3(k+1) - (k+9)] = 0

    Simplify inside the brackets:

    3(k+1) - (k+9) = 3k + 3 - k - 9 = 2k - 6

    Hence, the equation becomes:

    12(k+1)(2k - 6) = 0

    Setting each factor to zero:

    12(k+1) = 0 \quad \text{or} \quad 2k - 6 = 0

    Solve for ( k ):

    k + 1 = 0 \quad \Rightarrow \quad k = -1
    2k - 6 = 0 \quad \Rightarrow \quad 2k = 6 \quad \Rightarrow \quad k = 3

  4. Verify the condition ( k \neq -1 ):

    Since ( k = -1 ) is not allowed, ( k ) must be:

    k = 3

Final Answer:
The value of ( k ) for which the quadratic equation ( (k+1)x^2 - 6(k+1)x + 3(k+9) = 0 ) has equal roots is \boxed{3}