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Question: Find the roots of the quadratic equation \(\sqrt{2}{{x}^{2}}+7x+5\sqrt{2}=0\)....

Find the roots of the quadratic equation 2x2+7x+52=0\sqrt{2}{{x}^{2}}+7x+5\sqrt{2}=0.

Explanation

Solution

Hint: Here, in this given question, we can use the quadratic formula for the roots of the general quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0, which is x=b±b24ac2ax=\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 2\sqrt{2}, 7 and 525\sqrt{2} respectively and we will obtain the required roots of the given quadratic equation 2x2+7x+52=0\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 2x2+7x+52=0\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 ax2+bx+c=0a{{x}^{2}}+bx+c=0, which is x=b±b24ac2ax=\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 2\sqrt{2}, 7 and 525\sqrt{2} respectively.
So, putting these values in equation 1.2, we get,
x=7±724×2×522×2...........(1.3)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,
x=7±724×2×522×2 =7±494022 =7±922 =7±322...............(1.4) \begin{aligned} & 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{aligned}
Now, multiplying 2\sqrt{2} in both the numerator and denominator to rationalize the fraction, we get,
x=7±322=(7±3)×222×2=(7±3)24............(1.5)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
x=(7±3)24 =(7+3)24 or (73)24 =424 or 1024 =2 or 522 \begin{aligned} & 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{aligned}
Hence, we have obtained the values of x as 2-\sqrt{2} or 522\dfrac{-5\sqrt{2}}{2}.
Therefore, the roots of the equation 2x2+7x+52=0\sqrt{2}{{x}^{2}}+7x+5\sqrt{2}=0 are 2-\sqrt{2} and 522\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.