Question

How do you solve $\sqrt{2x-3}=2$?

Answer 1

Hint: Now to solve the given linear equation we will first eliminate the square root by squaring the terms on both sides. Then we have a simple linear equation in one variable. Hence we will rearrange the term and divide with the coefficient of x to find the value of x. Hence we have the solution to the given equation.

Complete step by step solution:
Now let us consider the given equation $\sqrt{2x-3}=2$.
This is a linear equation in one variable. We want to first bring this equation in its general form that is ax + b = 0 or ax = b.
Since we have a square root in the given equation we will first try to eliminate it by squaring.
Hence on squaring on both sides we get,
$\Rightarrow {{\left( \sqrt{2x-3} \right)}^{2}}={{2}^{2}}$
Now we know that ${{\left( \sqrt{x} \right)}^{2}}=x$ and ${{2}^{2}}=4$ . Hence using this we get,
$\Rightarrow 2x-3=4$
Now transposing 3 on RHS we get,
$\Rightarrow 2x=4+3$
$\Rightarrow 2x=7$
Now let us divide the whole equation by 2. Hence we get,
$\Rightarrow x=\dfrac{7}{2}$ .
Now we can see that the fraction is already in simplest form.
Hence we have the solution of the given equation is $x=\dfrac{7}{2}$ .

Note: Note that whenever we have square root in the equation we will first separate all the constant and root free terms with the root terms. Then we will square the equation to get rid of square root. After finding the solution for the equation, substitute the value of x and check if the equation holds. Hence we can check if the solution obtained is correct or not.