Question

Find the values of k for which the quadratic equation \[\left( {3k + 1} \right){x^2} + 2\left( {k + 1} \right)x + 1 = 0\] has equal roots. Also, find the roots.

Answer 1

Hint: First find the condition on the equation for it to have two equal roots. Verify the condition. Now substitute the given equations values into condition you get a quadratic in the variable and use the factorization method to solve first the variable. The value of the variable is required to result in the question.

Complete step-by-step answer:
Factorization: For factoring a quadratic follow the steps:
Allot \[{x^2}\] coefficient as “a”, x coefficient as “b”, constant as “c”. Find the product of the 2 numbers a, c. Let it be k. Find 2 numbers such that their product is k, sum is b. Rewrite the term bx in terms of those 2 numbers.
Now factor the first two terms.
Next factor the last two terms.
If we do it correctly our 2 new terms will have one more common factor.
Now take that common to get quadratic as a product of 2 terms, from which you get the roots.
Given quadratic equation in the question is written as follows:
\[\left( {3k + 1} \right){x^2} + 2\left( {k + 1} \right)x + 1 = 0\]
By comparing the equation to \[a{x^2} + bx + c\], we get values as,
The value of a in terms of k can be written as: \[a = 3k + 1\]
The value of b in terms of k can be written as: \[b = 2\left[ {k + 1} \right]\]
The value of c from the comparison can be written as: \[c = 1\]
By substituting these values into the condition of equal roots, we get:
\[{b^2} - 4ac = 0\]
We can convert the equation in terms of k, by writing as follows:
\[{\left( {2\left( {k + 1} \right)} \right)^2} - 4\left( {3k + 1} \right)\left( 1 \right) = 0\]
By simplifying the above equation we can write it as:
\[{\left[ {2k + 2} \right]^2} - 12k - 4 = 0\]
By using\[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\], we can write the above equation as:
\[4{k^2} + 4 + 8k - 12k - 4 = 0\]
By simplifying, we get the k equation as: \[4{k^2} - 4k = 0\]
By taking 4k as common, we get it as: \[4k\left( {k - 1} \right) = 0\]So, by this we get k values as \[k = 0,{\rm{ k = 1}}\]
By substituting \[k = 0{\rm{ or k = 1}}\], we get 2 equations as follows:
\[{x^2} + 2x + 1 = 0{\rm{ or 4}}{{\rm{x}}^{\rm{2}}}{\rm{ + 4x + 1 = 0}}\]
By simplifying, we can write the equation as follows:
\[{\left( {x + 1} \right)^2} = 0{\rm{ or }}{\left( {2x + 1} \right)^2} = 0\]
By this we get roots as\[x = - 1, - 1{\rm{ and x = }}\dfrac{{ - 1}}{2},\dfrac{{ - 1}}{2}\].
Therefore for \[k = 0,\]we get \[x = - 1, - 1\]and for\[k = 1\], we get \[x = \dfrac{{ - 1}}{2},\dfrac{{ - 1}}{2}.\]

Note: Be careful while verifying the condition for equal roots because that is the main point of solution. The idea of converting equation into square of term after substituting values of k is also important as it has equal roots anyhow you can convert it into square of a term.