Question

How do you solve $4{{x}^{2}}-16=0$?

Answer 1

We have to solve the problem using a quadratic formula. First we have to substitute the coefficients in the quadratic formula. Next we have to simplify it to get the desired results. Here we will consider the middle term as zero because it is not present in the given equation.

Complete step by step answer:
Given equation is
$4{{x}^{2}}-16=0$
The quadratic formula for the standard equation is $a{{x}^{2}}-bx+c=0$ is
$y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So using our equation we have to find $a,b,c$ values.
We will get values as

& a=4 \\\ & b=0 \\\ & c=-16 \\\ \end{aligned}$$ Now we have to substitute them in the formula we have $$\Rightarrow \dfrac{-0\pm \sqrt{{{0}^{2}}-4\times 4\times -16}}{2\times 4}$$ First we have to solve the part present inside the square root. We will get $$\Rightarrow \dfrac{0\pm \sqrt{0+16\times 16}}{8}$$ $$\Rightarrow \dfrac{0\pm \sqrt{256}}{8}$$ Now we have to find the square root of $$256$$.we will get $$\Rightarrow \dfrac{0\pm 16}{8}$$ Now we have to split it into two terms. We will get $$\Rightarrow \dfrac{0+16}{8}$$ $$\Rightarrow \dfrac{0-16}{8}$$ Now we have to do simple addition and subtraction to find the values. We will get $$\Rightarrow 2$$ $$\Rightarrow -2$$ So the $$x$$values we will get are $$x=2,x=-2$$. **Note:** We can solve this in another way also. We can also do this in a simple way by transferring $$16$$ to the RHS side. We will get $$\Rightarrow 4{{x}^{2}}=16$$ Now we have to divide both sides of the equation with $$4$$. $$\Rightarrow \dfrac{4{{x}^{2}}}{4}=\dfrac{16}{4}$$ By simplifying it we will get $$\Rightarrow {{x}^{2}}=4$$ Now we have apply square root on both side of equation $$\Rightarrow \sqrt{{{x}^{2}}}=\sqrt{4}$$ By simplifying it we will get $$\Rightarrow x=\pm 2$$ We can use any of the two methods that are discussed above. We cannot use factorization methods because we do not have a middle term to split. It is better to use a quadratic formula to solve this type of question.