
Solve for x the given quadratic polynomial \[4{{x}^{2}}+14x+6\]?
Answer
516.6k+ views
Hint: In this problem we have to solve the given quadratic equation and find the value of x. We know that to solve a quadratic equation, we can use two methods, the one is the quadratic formula method and the other is the factorisation method. To solve this by using the quadratic formula method, we have to use the quadratic formula to find the value of x. We know that the quadratic formula is \[x=\dfrac{-b\pm \sqrt{{{\left( b \right)}^{2}}-4\times a\times \left( c \right)}}{2a}\], we have to find a, b, c and substitute in the formula to get the va;ue of x.
Complete step-by-step solution:
We know that the given quadratic equation is,
\[4{{x}^{2}}+14x+6\] ….. (1)
We also know that a quadratic equation in standard form is,
\[a{{x}^{2}}+bx+c=0\] ……. (2)
We can now compare the two equations (1) and (2), we get
a = 4, b = 14, c = 6.
We know that the quadratic formula for the standard form \[a{{x}^{2}}+bx+c=0\] is
\[x=\dfrac{-b\pm \sqrt{{{\left( b \right)}^{2}}-4\times a\times \left( c \right)}}{2a}\]
Now we can substitute the value of a, b, c in the above formula, we get
\[\Rightarrow x=\dfrac{-\left( 14 \right)\pm \sqrt{{{\left( 14 \right)}^{2}}-4\times 4\times \left( 6 \right)}}{2\times 4}\]
Now we can simplify the above step, we get
\[\begin{align}
& \Rightarrow x=\dfrac{\left( -14 \right)\pm \sqrt{196-96}}{2\times 4} \\
& \Rightarrow x=\dfrac{-14\pm \sqrt{100}}{8}=\dfrac{-14\pm 10}{8} \\
\end{align}\]
Now we can separate the terms to simplify it,
\[\begin{align}
& \Rightarrow x=\dfrac{-14}{8}+\dfrac{10}{8}=\dfrac{-4}{8}=-\dfrac{1}{2} \\
& \Rightarrow x=\dfrac{-14}{8}-\dfrac{10}{8}=-\dfrac{24}{8}=-3 \\
\end{align}\]
Therefore, the value of \[x=-\dfrac{1}{2}\] and \[x=-3\].
Note: We can also use a simple factorisation method to solve this problem. Students may make mistakes in the quadratic formula part, which should be concentrated. we should also concentrate while substituting the values in the quadratic formula \[x=\dfrac{-b\pm \sqrt{{{\left( b \right)}^{2}}-4\times a\times \left( c \right)}}{2a}\] by finding the value of a, b, c.
Complete step-by-step solution:
We know that the given quadratic equation is,
\[4{{x}^{2}}+14x+6\] ….. (1)
We also know that a quadratic equation in standard form is,
\[a{{x}^{2}}+bx+c=0\] ……. (2)
We can now compare the two equations (1) and (2), we get
a = 4, b = 14, c = 6.
We know that the quadratic formula for the standard form \[a{{x}^{2}}+bx+c=0\] is
\[x=\dfrac{-b\pm \sqrt{{{\left( b \right)}^{2}}-4\times a\times \left( c \right)}}{2a}\]
Now we can substitute the value of a, b, c in the above formula, we get
\[\Rightarrow x=\dfrac{-\left( 14 \right)\pm \sqrt{{{\left( 14 \right)}^{2}}-4\times 4\times \left( 6 \right)}}{2\times 4}\]
Now we can simplify the above step, we get
\[\begin{align}
& \Rightarrow x=\dfrac{\left( -14 \right)\pm \sqrt{196-96}}{2\times 4} \\
& \Rightarrow x=\dfrac{-14\pm \sqrt{100}}{8}=\dfrac{-14\pm 10}{8} \\
\end{align}\]
Now we can separate the terms to simplify it,
\[\begin{align}
& \Rightarrow x=\dfrac{-14}{8}+\dfrac{10}{8}=\dfrac{-4}{8}=-\dfrac{1}{2} \\
& \Rightarrow x=\dfrac{-14}{8}-\dfrac{10}{8}=-\dfrac{24}{8}=-3 \\
\end{align}\]
Therefore, the value of \[x=-\dfrac{1}{2}\] and \[x=-3\].
Note: We can also use a simple factorisation method to solve this problem. Students may make mistakes in the quadratic formula part, which should be concentrated. we should also concentrate while substituting the values in the quadratic formula \[x=\dfrac{-b\pm \sqrt{{{\left( b \right)}^{2}}-4\times a\times \left( c \right)}}{2a}\] by finding the value of a, b, c.
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