Question

How do you solve \[3{x^2} - 5 = 0\]?

Answer 1

Hint: The given equation is a quadratic equation and so it will have two roots. It has no term containing \[x\]. So, to solve it we will shift the constant term to RHS and then take the square root of both sides to get the values of \[x\].

Complete step by step answer:
The given equation is:
\[3{x^2} - 5 = 0\]
Now we will shift the constant term that is \[5\] to the RHS. So, we have;
\[ \Rightarrow 3{x^2} = 5\]
Dividing both sides by \[3\], we get;
\[ \Rightarrow {x^2} = \dfrac{5}{3}\]
Now we will take the square root of both sides. So, we have;
\[ \Rightarrow \sqrt {{x^2}} = \sqrt {\dfrac{5}{3}} \]
Solving by using \[\sqrt {{x^2}} = |x|\]. So, we get;
\[ \Rightarrow |x| = \sqrt {\dfrac{5}{3}} \]
Removing the modulus, we get;
\[ \Rightarrow x = \pm \sqrt {\dfrac{5}{3}} \]

Note:
We can also solve the equation by finding the discriminant and then using the formula for the root of the quadratic equation.
The given equation is;
\[3{x^2} - 5 = 0\]
Comparing it with the standard form that is \[a{x^2} + bx + c\], we have \[a = 3,b = 0,c = - 5\]
Now we know that the discriminant of a quadratic equation is given by \[D = {b^2} - 4ac\]. So, putting the values we get;
\[ \Rightarrow D = {0^2} - 4 \times 3 \times \left( { - 5} \right)\]
Solving we get;
\[ \Rightarrow D = 60\]
As we can see that the discriminant is positive, the given quadratic equation will have real roots.
Now we know that the root is given by the formula;
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
We can write the term in the square root as discriminant. So, we get;
\[ \Rightarrow x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\]
Now putting the values, we get;
\[ \Rightarrow x = \dfrac{{ - 0 \pm \sqrt {60} }}{{2 \times 3}}\]
We can write it as;
\[ \Rightarrow x = \dfrac{{ \pm \sqrt {60} }}{6}\]
Now we will shift \[6\] under the square root sign. So, we have;
\[ \Rightarrow x = \dfrac{{ \pm \sqrt {60} }}{{\sqrt {36} }}\]
Now we will take the numerator and denominator under a single square root sign. So, we have;
\[ \Rightarrow x = \pm \sqrt {\dfrac{{60}}{{36}}} \]
On reducing it we get;
\[ \Rightarrow x = \pm \sqrt {\dfrac{5}{3}} \]