Find the roots of the quadratic equation $\sqrt{2}{{x}^{2}}+7x+5\sqrt{2}=0$.
Answer
636.3k+ views
Hint: Here, in this given question, we can use the quadratic formula for the roots of the general quadratic equation $a{{x}^{2}}+bx+c=0$, which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$in order to find the roots of the given equation. Just we have to substitute the values of a, b and c with $\sqrt{2}$, 7 and $5\sqrt{2}$ respectively and we will obtain the required roots of the given quadratic equation $\sqrt{2}{{x}^{2}}+7x+5\sqrt{2}=0$.
Complete stop-by-stop answer:
In this given question, we are asked to find out the roots of the given quadratic equation $\sqrt{2}{{x}^{2}}+7x+5\sqrt{2}=0$…………..(1.1)
A quadratic equation is an equation with the highest power of x as 2 or whose degree is 2.
Now, we are going to use the quadratic formula for the roots of the general quadratic equation $a{{x}^{2}}+bx+c=0$, which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$………………(1.2), in order to find the roots of the given equation.
In equation 1.1, the values of a, b and c are $\sqrt{2}$, 7 and $5\sqrt{2}$ respectively.
So, putting these values in equation 1.2, we get,
$x=\dfrac{-7\pm \sqrt{{{7}^{2}}-4\times \sqrt{2}\times 5\sqrt{2}}}{2\times \sqrt{2}}...........(1.3)$
Now, simplifying equation 1.3, we get,
$\begin{align}
& x=\dfrac{-7\pm \sqrt{{{7}^{2}}-4\times \sqrt{2}\times 5\sqrt{2}}}{2\times \sqrt{2}} \\
& =\dfrac{-7\pm \sqrt{49-40}}{2\sqrt{2}} \\
& =\dfrac{-7\pm \sqrt{9}}{2\sqrt{2}} \\
& =\dfrac{-7\pm 3}{2\sqrt{2}}...............(1.4) \\
\end{align}$
Now, multiplying $\sqrt{2}$ in both the numerator and denominator to rationalize the fraction, we get,
$x=\dfrac{-7\pm 3}{2\sqrt{2}}=\dfrac{\left( -7\pm 3 \right)\times \sqrt{2}}{2\sqrt{2}\times \sqrt{2}}=\dfrac{\left( -7\pm 3 \right)\sqrt{2}}{4}............(1.5)$
From equation 1.5, we get the values of x as
$\begin{align}
& x=\dfrac{\left( -7\pm 3 \right)\sqrt{2}}{4} \\
& =\dfrac{\left( -7+3 \right)\sqrt{2}}{4}\text{ or }\dfrac{\left( -7-3 \right)\sqrt{2}}{4} \\
& =\dfrac{-4\sqrt{2}}{4}\text{ or }\dfrac{-10\sqrt{2}}{4} \\
& =-\sqrt{2}\text{ or }\dfrac{-5\sqrt{2}}{2} \\
\end{align}$
Hence, we have obtained the values of x as \[-\sqrt{2}\] or \[\dfrac{-5\sqrt{2}}{2}\].
Therefore, the roots of the equation $\sqrt{2}{{x}^{2}}+7x+5\sqrt{2}=0$ are \[-\sqrt{2}\] and \[\dfrac{-5\sqrt{2}}{2}\].
Note: Although we have used the quadratic formula to find the roots of the given equation, we must note that there are also methods like factorization, completing the square and graphical method to solve this sort of question. We can use any of the methods, however the answer will remain the same.
Complete stop-by-stop answer:
In this given question, we are asked to find out the roots of the given quadratic equation $\sqrt{2}{{x}^{2}}+7x+5\sqrt{2}=0$…………..(1.1)
A quadratic equation is an equation with the highest power of x as 2 or whose degree is 2.
Now, we are going to use the quadratic formula for the roots of the general quadratic equation $a{{x}^{2}}+bx+c=0$, which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$………………(1.2), in order to find the roots of the given equation.
In equation 1.1, the values of a, b and c are $\sqrt{2}$, 7 and $5\sqrt{2}$ respectively.
So, putting these values in equation 1.2, we get,
$x=\dfrac{-7\pm \sqrt{{{7}^{2}}-4\times \sqrt{2}\times 5\sqrt{2}}}{2\times \sqrt{2}}...........(1.3)$
Now, simplifying equation 1.3, we get,
$\begin{align}
& x=\dfrac{-7\pm \sqrt{{{7}^{2}}-4\times \sqrt{2}\times 5\sqrt{2}}}{2\times \sqrt{2}} \\
& =\dfrac{-7\pm \sqrt{49-40}}{2\sqrt{2}} \\
& =\dfrac{-7\pm \sqrt{9}}{2\sqrt{2}} \\
& =\dfrac{-7\pm 3}{2\sqrt{2}}...............(1.4) \\
\end{align}$
Now, multiplying $\sqrt{2}$ in both the numerator and denominator to rationalize the fraction, we get,
$x=\dfrac{-7\pm 3}{2\sqrt{2}}=\dfrac{\left( -7\pm 3 \right)\times \sqrt{2}}{2\sqrt{2}\times \sqrt{2}}=\dfrac{\left( -7\pm 3 \right)\sqrt{2}}{4}............(1.5)$
From equation 1.5, we get the values of x as
$\begin{align}
& x=\dfrac{\left( -7\pm 3 \right)\sqrt{2}}{4} \\
& =\dfrac{\left( -7+3 \right)\sqrt{2}}{4}\text{ or }\dfrac{\left( -7-3 \right)\sqrt{2}}{4} \\
& =\dfrac{-4\sqrt{2}}{4}\text{ or }\dfrac{-10\sqrt{2}}{4} \\
& =-\sqrt{2}\text{ or }\dfrac{-5\sqrt{2}}{2} \\
\end{align}$
Hence, we have obtained the values of x as \[-\sqrt{2}\] or \[\dfrac{-5\sqrt{2}}{2}\].
Therefore, the roots of the equation $\sqrt{2}{{x}^{2}}+7x+5\sqrt{2}=0$ are \[-\sqrt{2}\] and \[\dfrac{-5\sqrt{2}}{2}\].
Note: Although we have used the quadratic formula to find the roots of the given equation, we must note that there are also methods like factorization, completing the square and graphical method to solve this sort of question. We can use any of the methods, however the answer will remain the same.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

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

Difference Between Prokaryotic Cells and Eukaryotic Cells

Two of the body parts which do not appear in MRI are class 11 biology CBSE

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

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

10 examples of friction in our daily life

