
If -5 is a root of the quadratic equation \[2{{x}^{2}}+px-15=0\] and the quadratic equation \[p\left( {{x}^{2}}+x \right)+k=0\] has equal roots, find the value of k.
Answer
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Hint: In this problem, we have to find the value of k, where : If -5 is a root of the quadratic equation \[2{{x}^{2}}+px-15=0\] and the quadratic equation \[p\left( {{x}^{2}}+x \right)+k=0\] has equal roots. We can first take the first equation, where we are given the root of it, so we can substitute the root value in the equation and get the value of p, we can then substitute the p value in the second equation and the root value, as it has same root value, to get the value of k.
Complete step by step answer:
Here we have to find the value of k for the given equation.
We are given that -5 is a root of the quadratic equation \[2{{x}^{2}}+px-15=0\] and the quadratic equation \[p\left( {{x}^{2}}+x \right)+k=0\] has equal roots.
We can now find the value of p.
We know that the root of the equation \[2{{x}^{2}}+px-15=0\] is -5, we can now write it as,
\[\Rightarrow 2{{\left( -5 \right)}^{2}}+p\left( -5 \right)-15=0\]
We can now simplify the above step, we get
\[\begin{align}
& \Rightarrow 50-5p-15=0 \\
& \Rightarrow -5p=-35 \\
& \Rightarrow p=7 \\
\end{align}\]
The value of p = 7.
We can now substitute the p value in \[p\left( {{x}^{2}}+x \right)+k=0\].
\[\Rightarrow 7\left( {{x}^{2}}+x \right)+k=0\]
We are given that the root value is same for the above equation, so we get
\[\Rightarrow 7\left( {{\left( -5 \right)}^{2}}-5 \right)+k=0\]
We can now simplify the above step, we get
\[\begin{align}
& \Rightarrow 7\left( 25-5 \right)+k=0 \\
& \Rightarrow 140+k=0 \\
& \Rightarrow k=-140 \\
\end{align}\]
Therefore, the value of k is -140.
Note: Students make mistakes while substituting the value for x, as we are given a root value, which is the value of x. We should also note that the given both equations have the same root, i.e. x = -5. We can first find the value of p in the first equation to substitute it in the second equation to get the value of k.
Complete step by step answer:
Here we have to find the value of k for the given equation.
We are given that -5 is a root of the quadratic equation \[2{{x}^{2}}+px-15=0\] and the quadratic equation \[p\left( {{x}^{2}}+x \right)+k=0\] has equal roots.
We can now find the value of p.
We know that the root of the equation \[2{{x}^{2}}+px-15=0\] is -5, we can now write it as,
\[\Rightarrow 2{{\left( -5 \right)}^{2}}+p\left( -5 \right)-15=0\]
We can now simplify the above step, we get
\[\begin{align}
& \Rightarrow 50-5p-15=0 \\
& \Rightarrow -5p=-35 \\
& \Rightarrow p=7 \\
\end{align}\]
The value of p = 7.
We can now substitute the p value in \[p\left( {{x}^{2}}+x \right)+k=0\].
\[\Rightarrow 7\left( {{x}^{2}}+x \right)+k=0\]
We are given that the root value is same for the above equation, so we get
\[\Rightarrow 7\left( {{\left( -5 \right)}^{2}}-5 \right)+k=0\]
We can now simplify the above step, we get
\[\begin{align}
& \Rightarrow 7\left( 25-5 \right)+k=0 \\
& \Rightarrow 140+k=0 \\
& \Rightarrow k=-140 \\
\end{align}\]
Therefore, the value of k is -140.
Note: Students make mistakes while substituting the value for x, as we are given a root value, which is the value of x. We should also note that the given both equations have the same root, i.e. x = -5. We can first find the value of p in the first equation to substitute it in the second equation to get the value of k.
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