How do you solve the quadratic equation using the quadratic formula, given $9{{x}^{2}}-11=6x$?
Answer
630.6k+ views
Hint: We will rearrange the given equation in its standard form. We will compare the given equation with the general quadratic equation and equate the corresponding coefficients. Then we will see the quadratic formula for the solution of a general quadratic equation. We will substitute corresponding coefficients from the given equation in the quadratic formula and obtain the solution.
Complete step by step answer:
The given equation is $9{{x}^{2}}-11=6x$. Rearranging this equation in its standard form, we get
$9{{x}^{2}}-6x-11=0$
We know that the general quadratic equation is given as the following,
$a{{x}^{2}}+bx+c=0$
Now, comparing the given quadratic equation with the general quadratic equation, we get the following as corresponding coefficients,
$a=9$, $b=-6$ and $c=-11$.
The quadratic formula for obtaining the solution of a general quadratic equation is given by
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Next, we will substitute the corresponding coefficients from the given quadratic equation in the above formula. Doing this, we get
$x=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 9 \right)\left( -11 \right)}}{2\left( 9 \right)}$
Simplifying the above equation, we get
$\begin{align}
& x=\dfrac{6\pm \sqrt{36+396}}{18} \\
& \therefore x=\dfrac{6\pm \sqrt{432}}{18} \\
\end{align}$
Now, we can see that 432 has 2 in it units place and therefore it is not a perfect square. Upon factorizing 432, we can see that
$432=12\times 12\times 3$
Therefore, we have $\sqrt{432}=12\sqrt{3}$. Substituting this in the above equation we get,
$x=\dfrac{6\pm 12\sqrt{3}}{18}$
Cancelling out the factor 6 from the numerator and the denominator, we get
$x=\dfrac{1\pm 2\sqrt{3}}{3}$
Therefore, the solution of the given quadratic equation is $x=\dfrac{1+2\sqrt{3}}{3}$ and $x=\dfrac{1-2\sqrt{3}}{3}$.
Note:
We should be familiar with the fact that the number having the digits 2, 3, 7 or 8 in the unit’ place cannot be perfect squares. Apart from the quadratic formula, there are other methods of solving a quadratic equation. These methods are factorization and completing the square method. When given a choice for the method, we should choose according to our convenience and ease of calculation.
Complete step by step answer:
The given equation is $9{{x}^{2}}-11=6x$. Rearranging this equation in its standard form, we get
$9{{x}^{2}}-6x-11=0$
We know that the general quadratic equation is given as the following,
$a{{x}^{2}}+bx+c=0$
Now, comparing the given quadratic equation with the general quadratic equation, we get the following as corresponding coefficients,
$a=9$, $b=-6$ and $c=-11$.
The quadratic formula for obtaining the solution of a general quadratic equation is given by
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Next, we will substitute the corresponding coefficients from the given quadratic equation in the above formula. Doing this, we get
$x=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 9 \right)\left( -11 \right)}}{2\left( 9 \right)}$
Simplifying the above equation, we get
$\begin{align}
& x=\dfrac{6\pm \sqrt{36+396}}{18} \\
& \therefore x=\dfrac{6\pm \sqrt{432}}{18} \\
\end{align}$
Now, we can see that 432 has 2 in it units place and therefore it is not a perfect square. Upon factorizing 432, we can see that
$432=12\times 12\times 3$
Therefore, we have $\sqrt{432}=12\sqrt{3}$. Substituting this in the above equation we get,
$x=\dfrac{6\pm 12\sqrt{3}}{18}$
Cancelling out the factor 6 from the numerator and the denominator, we get
$x=\dfrac{1\pm 2\sqrt{3}}{3}$
Therefore, the solution of the given quadratic equation is $x=\dfrac{1+2\sqrt{3}}{3}$ and $x=\dfrac{1-2\sqrt{3}}{3}$.
Note:
We should be familiar with the fact that the number having the digits 2, 3, 7 or 8 in the unit’ place cannot be perfect squares. Apart from the quadratic formula, there are other methods of solving a quadratic equation. These methods are factorization and completing the square method. When given a choice for the method, we should choose according to our convenience and ease of calculation.
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