Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

How do you solve $4{{x}^{2}}=28$?

Answer
VerifiedVerified
491.1k+ views
Hint: We have been given a quadratic equation in one variable. However, in the equation, the 1-degree term in x-variable is missing and only the 2-degree term and the constant term is present. Therefore, we will not be using the conventional factoring method of solving the quadratic equations. Rather, we will first find the discriminant of the quadratic equation and then further find the value of x.


Complete step-by-step solution:
A quadratic equation is given in its standard form as $a{{x}^{2}}+bx+c=0$. In this form, both the 2-degree and 1-degree terms are assigned the coefficients $a$ and $b$ respectively.
We know that the discriminant, $D$ of any quadratic equation is given as:
$D={{b}^{2}}-4ac$
And then the solutions of the quadratic equation are expressed as:
$x=\dfrac{-b\pm \sqrt{D}}{2a}$
We are given the equation, $4{{x}^{2}}=28$.
We shall transpose 28 to the left-hand side to convert the quadratic equation in the standard form.
$\Rightarrow 4{{x}^{2}}-28=0$
Here, on comparing the coefficients, we have $a=4$, $b=0$ and $c=-28$.
Applying the formula of discriminant and substituting all values, we get
$\Rightarrow D={{0}^{2}}-4\left( 4 \right)\left( -28 \right)$
$\Rightarrow D=448$
Now, we shall finally find the roots of the quadratic equation by putting the value of discriminant calculated.
$x=\dfrac{-0\pm \sqrt{448}}{2\left( 4 \right)}$
$\Rightarrow x=\dfrac{\pm \sqrt{448}}{8}$
 On factoring 448, we find that $448={{4}^{2}}{{.2}^{2}}.7$ . Therefore, $\sqrt{448}= 8\sqrt{7}$
$\Rightarrow x=\dfrac{\pm 8\sqrt{7}}{8}$
Cancelling 8 from numerator and denominator, we get
$\Rightarrow x=\pm \sqrt{7}$
Therefore, the solutions of $4{{x}^{2}}=28$ are $x=\sqrt{7}$ and $x=-\sqrt{7}$.

Note: Another method of solving this problem was by dividing the entire equation by 4 and then finding the square root of the constant to calculate x. Although this could have been an easier approach but it is not recommended it involves basic algebra and does not include any properties related to the quadratic equations.