If the equation ${x^2} + kx + k = 0$ has only one solution, find the possible values of k.
Answer
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Hint – In order to solve this equation, we have to find out the nature and roots of a quadratic expression by using discrimination method.
Complete step-by-step answer:
It is given that,
${x^2} + kx + k = 0$
Compare above equation with,
$a{x^2} + bx + c = 0$ , we get
Where, $a = 1,{\text{ b = k, c = k}}$
If the equation has real and equal roots, then
Discriminant (D)=0
$
\Rightarrow {b^2} - 4ac \\
\Rightarrow {k^2} - 4 \times 1 \times k = 0 \\
\Rightarrow k\left( {k - 4} \right) = 0 \\
$
$ \Rightarrow k = 0{\text{ or k - 4 = 0}}$
$ \Rightarrow k = 0{\text{ or }}k{\text{ = 4 }}$
Therefore,
$k = 0{\text{ or }}k = 4$.
Note – In this particular question, we have used the discrimination method that is ${b^2} - 4ac{\text{ = 0}}$to get our required solution. There are so many methods by which one can solve this question, one method to solve this question is by using the squares method to get equal or double roots.
Complete step-by-step answer:
It is given that,
${x^2} + kx + k = 0$
Compare above equation with,
$a{x^2} + bx + c = 0$ , we get
Where, $a = 1,{\text{ b = k, c = k}}$
If the equation has real and equal roots, then
Discriminant (D)=0
$
\Rightarrow {b^2} - 4ac \\
\Rightarrow {k^2} - 4 \times 1 \times k = 0 \\
\Rightarrow k\left( {k - 4} \right) = 0 \\
$
$ \Rightarrow k = 0{\text{ or k - 4 = 0}}$
$ \Rightarrow k = 0{\text{ or }}k{\text{ = 4 }}$
Therefore,
$k = 0{\text{ or }}k = 4$.
Note – In this particular question, we have used the discrimination method that is ${b^2} - 4ac{\text{ = 0}}$to get our required solution. There are so many methods by which one can solve this question, one method to solve this question is by using the squares method to get equal or double roots.
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