Question

How do you solve $2{x^2} + 12x + 9 = 0?$

Answer 1

Hint: The given equation is of the form of a quadratic equation. Here we need to find the roots of the equation. i.e. we need to find the values of the variable x. Compare the given quadratic equation with the general quadratic equation and substitute the values in the formula of finding the roots of the equation.
If $a{x^2} + bx + c = 0$ is an general quadratic equation, then its roots are found using the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Complete step by step solution:
We are given the quadratic equation
$2{x^2} + 12x + 9 = 0$ ……(1)
If $a{x^2} + bx + c = 0$ is an general quadratic equation, then we find its roots using the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. ……(2)
On comparing with the general quadratic equation $a{x^2} + bx + c = 0$, we get,
$a = 2,$ $b = 12,$ $c = 9.$
Substitute the values of $a,$ $b$ and $c$ in the formula of finding the roots of the equation given by the equation (2), we get,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$ \Rightarrow x = \dfrac{{ - 12 \pm \sqrt {{{(12)}^2} - 4(2)(9)} }}{{2 \times 2}}$
Square the values inside the square root in the numerator of the fraction and then we obtain the solution as,
$ \Rightarrow x = \dfrac{{ - 12 \pm \sqrt {144 - 72} }}{4}$
Now calculate the value under the square root in the fraction we obtain,
$ \Rightarrow x = \dfrac{{ - 12 \pm \sqrt {72} }}{4}$ ……(3)
We now simplify $\sqrt {72} $.
We can write $\sqrt {72} = \sqrt {8 \times 9} $
                                   $ = \sqrt {4 \times 2 \times 9} $
We know that $\sqrt 4 = 2$ and $\sqrt 9 = 3$.
 Now substituting it we obtain $\sqrt {72} = 2 \times 3\sqrt 2 $
                          $\therefore \sqrt {72} $ $ = 6\sqrt 2 $
Hence the equation (3) becomes,
$ \Rightarrow x = \dfrac{{ - 12 \pm 6\sqrt 2 }}{4}$
Take the common factor 6 from the both terms in the numerator, we get,
$ \Rightarrow x = \dfrac{{6( - 2 \pm \sqrt 2 )}}{4}$
Now 4 in the denominator can be written as $4 = 2 \times 2$.
Hence we obtain,
$ \Rightarrow x = \dfrac{{6( - 2 \pm \sqrt 2 )}}{{2 \times 2}}$
Now cancelling the terms in the numerator and denominator we get,
$ \Rightarrow x = \dfrac{{3( - 2 \pm \sqrt 2 )}}{2}$
$ \Rightarrow x = \dfrac{1}{2}\left( { - 6 \pm 3\sqrt 2 } \right)$
$ \Rightarrow x = - 3 \pm \dfrac{{3\sqrt 2 }}{2}$
Hence we obtain the values if x as $x = - 3 + \dfrac{{3\sqrt 2 }}{2}$ and $x = - 3 - \dfrac{{3\sqrt 2 }}{2}$.
Therefore the roots of the quadratic equation $2{x^2} + 12x + 9 = 0$ is given by,
$x = - 3 + \dfrac{{3\sqrt 2 }}{2}$ and $x = - 3 - \dfrac{{3\sqrt 2 }}{2}$.

Note:
Here we can’t use the method of factorization to find the roots or values of x for this question. Most of the students write the coefficient of x as $3x + 4x$ or $6x + 2x$, but this is wrong as it will give us sum as $7x$ or $8x$, but it will not give us the product of terms to be coefficient of ${x^2}$. We need to keep in mind that the coefficient of x should be broken in such a way that its sum gives us the value of the coefficient x and its product gives the value of the coefficient of ${x^2}$.