
Find the roots of the quadratic equation $16{{x}^{2}}-24x-1=0$ by using the quadratic formula.
Answer
587.4k+ views
Hint: We will be using the quadratic formula to find the roots of the given quadratic equation. The general quadratic equation is of the form $a{{x}^{2}}+bx+c=0$, where $a\ne 0$. We know that we can obtain the roots of a quadratic equation by using the quadratic formula. The quadratic formula is given by
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Complete step-by-step solution:
The given quadratic equation is $16{{x}^{2}}-24x-1=0$. The general quadratic equation is $a{{x}^{2}}+bx+c=0$. Comparing the given quadratic equation with the general quadratic equation, we will get the values for $a$, $b$ and $c$. We have the following values for $a$, $b$ and $c$:
\[\begin{align}
& a=16 \\
& b=-24 \\
& c=-1 \\
\end{align}\]
The quadratic formula to obtain the roots of a general quadratic equation is given by
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
We have the formula and the values for $a$, $b$ and $c$. Now, we will substitute the values for $a$ , $b$ and $c$ in this formula. So we have the following equation,
\[x=\dfrac{-(-24)\pm \sqrt{{{(-24)}^{2}}-4\cdot (16)\cdot (-1)}}{2\cdot 16}\]
We will simplify the above equation to obtain the roots of the quadratic equation in the following manner,
\[x=\dfrac{24\pm \sqrt{576+64}}{32}\]
\[\begin{align}
& x=\dfrac{24\pm \sqrt{640}}{32} \\
& =\dfrac{24\pm 8\sqrt{10}}{32}
\end{align}\]
We can simplify the above equation by factoring the numerator. We get the following expression,
\[\begin{align}
& x=\dfrac{8\cdot (3\pm \sqrt{10})}{32} \\
& =\dfrac{3\pm \sqrt{10}}{4}
\end{align}\]
So we have obtained the two roots that have values $\dfrac{3+\sqrt{10}}{4}$ and $\dfrac{3-\sqrt{10}}{4}$ .
Note: The signs of the values of $a$ ,$b$ and $c$ must be carefully written. It is possible to misplace the signs while substituting the values in the quadratic formula. We can find the roots of a quadratic equation by other methods, such as factorization method, completing square method or drawing a graph.
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
Complete step-by-step solution:
The given quadratic equation is $16{{x}^{2}}-24x-1=0$. The general quadratic equation is $a{{x}^{2}}+bx+c=0$. Comparing the given quadratic equation with the general quadratic equation, we will get the values for $a$, $b$ and $c$. We have the following values for $a$, $b$ and $c$:
\[\begin{align}
& a=16 \\
& b=-24 \\
& c=-1 \\
\end{align}\]
The quadratic formula to obtain the roots of a general quadratic equation is given by
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
We have the formula and the values for $a$, $b$ and $c$. Now, we will substitute the values for $a$ , $b$ and $c$ in this formula. So we have the following equation,
\[x=\dfrac{-(-24)\pm \sqrt{{{(-24)}^{2}}-4\cdot (16)\cdot (-1)}}{2\cdot 16}\]
We will simplify the above equation to obtain the roots of the quadratic equation in the following manner,
\[x=\dfrac{24\pm \sqrt{576+64}}{32}\]
\[\begin{align}
& x=\dfrac{24\pm \sqrt{640}}{32} \\
& =\dfrac{24\pm 8\sqrt{10}}{32}
\end{align}\]
We can simplify the above equation by factoring the numerator. We get the following expression,
\[\begin{align}
& x=\dfrac{8\cdot (3\pm \sqrt{10})}{32} \\
& =\dfrac{3\pm \sqrt{10}}{4}
\end{align}\]
So we have obtained the two roots that have values $\dfrac{3+\sqrt{10}}{4}$ and $\dfrac{3-\sqrt{10}}{4}$ .
Note: The signs of the values of $a$ ,$b$ and $c$ must be carefully written. It is possible to misplace the signs while substituting the values in the quadratic formula. We can find the roots of a quadratic equation by other methods, such as factorization method, completing square method or drawing a graph.
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