Question

How do you solve $5{x^2} - 180 = 0?$

Answer 1

This question describes the operation of addition/ subtraction/ multiplication/ division. Also, we need to know the basic form of a quadratic equation and the formula to find out the term $x$ in the quadratic equation. In the quadratic equation if any one term is missing we can assume that as zero to make the easy calculation. Also, we need to know the square root values of basic numbers.

Complete step by step solution:
The given question is shown below,
$5{x^2} - 180 = 0?$
The above equation can also be written as,
$5{x^2} + 0x - 180 = 0 \to \left( 1 \right)$
We know that the basic form of a quadratic equation is,
$a{x^2} + bx + c = 0 \to \left( 2 \right)$
Then,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \to \left( 3 \right)$
By comparing the equation $\left( 1 \right)$and$\left( 2 \right)$, we get the values
of $a,b$and$c$.

$$\left( 1 \right) \to 5{x^2} + 0x - 180 = 0 \\\ \left( 2 \right) \to a{x^2} + bx + c = 0 \\\$$

So, we get the value of $a$is$5$, the value of$b$is$0$, and the value of $c$is$- 180$. Let’s substitute these values in the equation$\left( 3 \right)$, we get
$\left( 3 \right) \to x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$x = \dfrac{{ - \left( 0 \right) \pm \sqrt {{{\left( 0 \right)}^2} - 4 \times 5 \times - 180} }}{{2 \times 5}}$
$x = \dfrac{{0 \pm \sqrt {0 + 3600} }}{{10}}$
We know that, ${60^2} = 3600$.
So, the above equation can also be written as,
$x = \dfrac{{0 \pm \sqrt {{{60}^2}} }}{{10}}$
The square and square root are cancelled each other, so we get
$x = \dfrac{{0 \pm 60}}{{10}}$
Case: 1

$$x = \dfrac{{0 + 60}}{{10}} \\\ x = \dfrac{{ + 60}}{{10}} \\\ x = 6 \\\$$

Case: 2

$$x = \dfrac{{0 - 60}}{{10}} \\\ x = \dfrac{{ - 60}}{{10}} \\\ x = - 6 \\\$$

So, we get two answers as the value of $x$ which is given below,

$$x = 6 \\\ x = - 6 \\\$$

So, the final answer is,
$x = 6$and$x = - 6$

Note: This type of question involves the operation of addition/ subtraction/ multiplication/ division. Note that the denominator value would not be equal to zero. When ${n^2}$ is placed inside the square root we can cancel the square ${n^2}$ with the square root. Also, note that the value of $+ 0, - 0$and ${0^2}$ is always zero. If we have$\pm$in the equation, then we would find two values for $x$.