
Find the roots of the quadratic equation $16{{x}^{2}}-24x-1=0$ by using the quadratic formula.
Answer
526.2k+ 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.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE
