Question

How do you solve the given equation $20{(x - 3)^2} = 625$?

Answer 1

Hint: In order to solve this question we need to transform the all the values from algebraic to the other side keeping algebraic one in one side then by taking the square root of that value and by transformation of the remaining to that side we will get the final answer.

Complete Step by step answer:
For solving this question first we will be transform the value to the other side of the algebraic form:
$ \Rightarrow {(x - 3)^2} = \dfrac{{625}}{{20}}$
On further solving it we will get:
$ \Rightarrow {\left( {x - 3} \right)^2} = 31.25$
Taking the square root on both side;
$ \Rightarrow x - 3 = \pm \sqrt {31.25} $
Now taking the positive value
$ \Rightarrow x = \sqrt {31.25} + 3$
On further solving; x = 8.6
On taking the negative value;
$ \Rightarrow x = - \sqrt {31.24} + 3$
On further solving; x = -2.6
So the values of x are 8.6 and -2.6.

Note: There is another way of solving this equation and finding the values of x:-
$ \Rightarrow 20{\left( {x - 3} \right)^2} = 625$
Now squaring the term;
$ \Rightarrow 20({x^2} + 9 - 6x) = 625$
On further solving;
$ \Rightarrow 20{x^2} + 180 - 120x = 625$
Substituting the value 625 on another side and making it in such form so that it may be converted into applying the dharacharya formula;
$ \Rightarrow 20{x^2} - 120x - 445 = 0$
Dividing whole equation by 5 we get;
$ \Rightarrow 4{x^2} - 24x - 89 = 0$
Applying the dharacharya formula;
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
According to the question;
a = 4,b = -24,c = 89.
Putting all the values in this formula get; x=8.6, -2.6