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In the following, determine the value of k for which the given value is a solution of the equation
\[{x^2} + 2ax - k = 0,x = - a\] .

Answer
VerifiedVerified
578.1k+ views
Hint: If a number has any factor that means the number is completely divisible by its factor that is after division we will get remainder zero.
When there is a value mentioned and satisfies a given equation or value of any variable which satisfies that equation then the value of the equation after putting the value of variable becomes zero which means we consider it as a factor and by this value of other remaining terms or unknown variables is found.
Firstly we will put value of $x$ as given in question and solve by proceeding further we will get value of k
Then we will verify resulting value of k by putting it back into equation
Equate left hand side and right hand side.

Complete step-by-step answer:
We have
\[{x^2} + 2ax - k = 0\] ……(1)
now putting \[x = - a\] in equation (1) since it is a solution to equation (1) and equating it with zero
$\Rightarrow$${\left( { - a} \right)^2} + 2a \times - a - k = 0$
We know that
$\Rightarrow$\[{\left( { - a} \right)^2} = - a \times - a = {a^2}\]
Also we have,
$\Rightarrow$\[2a \times \left( { - a} \right) = - 2{a^2}\]
Therefore,
$\Rightarrow$\[{a^2} - 2{a^2}-k = 0\]
$\Rightarrow$\[ - {a^2} = k\]

Therefore, \[k = - {a^{2}}\]

Note: You can verify the value of k whether it is correct or not by substituting the value of k in the equation.
So let’s check
We have \[x = - a\] and \[k = - {a^2}\]
Substituting values again in equation for verification
\[{x^2} + 2ax - k = 0\]
$\Rightarrow$\[{a^2} - 2{a^2}-\left( { - {a^2}} \right) = 0\]
$\Rightarrow$\[{a^2} + {a^2} - 2{a^2} = 0\]
$\Rightarrow$$2{a^2} - 2{a^2} = 0$
$\Rightarrow$$0 = 0$
Therefore, we have LHS = RHS
So it confirms that we have calculated the correct value of k.