How do you find the value of the discriminant and state the type of the solutions given $4{{k}^{2}}+5k+4=-3k$?
Answer
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Hint: In this problem we need to calculate the discriminant value of the given quadrant equation. We know that the given discriminant of the quadrant equation $a{{x}^{2}}+bx+c=0$ is given by ${{b}^{2}}-4ac$. In this problem we can observe that the given equation is not in the form of the standard equation which is $a{{x}^{2}}+bx+c=0$. So, we will convert the given equation into a standard quadratic equation by applying some arithmetical operations. After getting the quadratic equation in standard from we will compare the obtained equation with $a{{x}^{2}}+bx+c=0$ and calculate the value of discriminate which is ${{b}^{2}}-4ac$. Based on the nature of the discriminant we can decide the type of solution of the given equation.
Complete step-by-step solution:
Given equation is $4{{k}^{2}}+5k+4=-3k$.
In the above equation we can observe that the $-3k$ in the RHS side. So, adding the $3k$ on both sides of the above equation, then we will get
$\Rightarrow 4{{k}^{2}}+5x+4+3k=-3k+3k$
Simplifying the above equation, then we will get
$\Rightarrow 4{{k}^{2}}+8k+4=0$
Comparing the above equation with standard equation $a{{x}^{2}}+bx+c=0$, then we will get
$a=4$, $b=8$, $c=4$.
Now the value of the discriminant of the given equation will be
$\begin{align}
& \Rightarrow {{b}^{2}}-4ac={{8}^{2}}-4\left( 4 \right)\left( 4 \right) \\
& \Rightarrow {{b}^{2}}-4ac=64-64 \\
& \Rightarrow {{b}^{2}}-4ac=0 \\
\end{align}$
The discriminant value of the given equation $4{{k}^{2}}+5k+4=-3k$ is $0$.
From the above discriminant value, we can say that the given equation $4{{k}^{2}}+5k+4=-3k$ has only one real repeated root.
Note: In this problem we have the discriminant value as zero, so we have written that the equation has one real and repeated root. If we get the discriminant value as less than zero or negative value, then the equation has two imaginary roots. If that discriminant value is greater than zero or positive value, then the equation has two distinct real roots.
Complete step-by-step solution:
Given equation is $4{{k}^{2}}+5k+4=-3k$.
In the above equation we can observe that the $-3k$ in the RHS side. So, adding the $3k$ on both sides of the above equation, then we will get
$\Rightarrow 4{{k}^{2}}+5x+4+3k=-3k+3k$
Simplifying the above equation, then we will get
$\Rightarrow 4{{k}^{2}}+8k+4=0$
Comparing the above equation with standard equation $a{{x}^{2}}+bx+c=0$, then we will get
$a=4$, $b=8$, $c=4$.
Now the value of the discriminant of the given equation will be
$\begin{align}
& \Rightarrow {{b}^{2}}-4ac={{8}^{2}}-4\left( 4 \right)\left( 4 \right) \\
& \Rightarrow {{b}^{2}}-4ac=64-64 \\
& \Rightarrow {{b}^{2}}-4ac=0 \\
\end{align}$
The discriminant value of the given equation $4{{k}^{2}}+5k+4=-3k$ is $0$.
From the above discriminant value, we can say that the given equation $4{{k}^{2}}+5k+4=-3k$ has only one real repeated root.
Note: In this problem we have the discriminant value as zero, so we have written that the equation has one real and repeated root. If we get the discriminant value as less than zero or negative value, then the equation has two imaginary roots. If that discriminant value is greater than zero or positive value, then the equation has two distinct real roots.
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