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# How do you solve $9{{x}^{2}}-16=0$ ?

Last updated date: 18th Jul 2024
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Hint: We will solve the above question by using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . First, we will put the values in the above equation from the given question according to the general quadratic equation. Then, I will find the values of x.

Let us solve the question.
We have given the equation as $9{{x}^{2}}-16=0$ .
As we know, the general form of the quadratic equation is $a{{x}^{2}}+bx+c=0$ .
And the values of x according to the equation $a{{x}^{2}}+bx+c=0$ are $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
According to the above equations, we can say that
a=9, b=0, and c=-16.
Now, we will plug them in the equation $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ , we get
$x=\dfrac{-0\pm \sqrt{{{0}^{2}}-4\times 9\times (-16)}}{2\times 9}$
Now, we will simplify the above equation.
$\Rightarrow x=\dfrac{\pm \sqrt{-4\times 9\times (-16)}}{2\times 9}$
$\Rightarrow x=\pm \dfrac{\sqrt{36\times 16}}{18}$
$\Rightarrow x=\pm \dfrac{24}{18}$
$\Rightarrow x=\pm \dfrac{4}{3}$
Hence, the values of x are $\dfrac{4}{3}$ and $-\dfrac{4}{3}$ .

Note:
In this type of question, we should have a proper knowledge in quadratic equations. We should be careful of the signs. We can solve this by a second method simply. Suppose, we have the equation$9{{x}^{2}}-16=0$
We will take the number 16 to the right side of the equation. After that, we will find the value of x.
$\Rightarrow {{x}^{2}}=\dfrac{16}{9}$
Now, by taking the square root both sides of the above equation, we get
$\Rightarrow x=\sqrt{\dfrac{16}{9}}$
$\Rightarrow x=\pm \dfrac{4}{3}$
Therefore, we get the values of x as $\dfrac{4}{3}$ and $-\dfrac{4}{3}$.
We have another method of solving the question.
We know that ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$ .
Using this formula, we can find the solution.
Therefore, we can say that $9{{x}^{2}}-16=(3x+4)(3x-4)$ .
Then, (3x+4)(3x-4)=0
Hence, we can say that $x=-\dfrac{4}{3}$ and $x=\dfrac{4}{3}$ .
As we can say that all above methods are giving satisfactory solutions. So, we can use any of the methods to solve them.