Question

Write standard form of quadratic equation and find the roots of the equation $3x^2+5\sqrt{2}x+2=0$ using general formula.

Answer 1

The standard form of a quadratic equation is $a{x}^2+bx+c=0,a\ne 0$.
(That is, when the terms of the quadratic equation are written in descending order of their degrees, then we get the standard form of the equation.)
For the equation $3x^2+5\sqrt{2}x+2=0$, we get
$a=3,b=5\sqrt{2},c=2$
Using the general form, we get
$D=b^2-4ac=(5\sqrt{2}{)}^2-4(3\left)\right(2)=26.$
Here, $D>0,$ so the equation has real and distinct roots.
Let the roots be $\alpha$ and $\beta$.
So, $\alpha =\frac{-a+\sqrt{D}}{2a}$ and $\beta =\frac{-b-\sqrt{D}}{2a}$

$\alpha =\frac{-5\sqrt{2}+\sqrt{26}}{2\left(3\right)}$ and $\beta =\frac{-5\sqrt{2}-\sqrt{26}}{2\left(3\right)}$

$\alpha =\frac{-5\sqrt{2}+\sqrt{26}}{6}$ and $\beta =\frac{-5\sqrt{2}-\sqrt{26}}{6}$
$\therefore$ Roots of the given equation are $\frac{-5\sqrt{2}+\sqrt{26}}{6}$ and $\frac{-5\sqrt{2}-\sqrt{26}}{6}$.