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Question: How do you solve \({\left( {x - 2} \right)^2} = 25\)?...

How do you solve (x2)2=25{\left( {x - 2} \right)^2} = 25?

Explanation

Solution

In order to solve the above expression transpose square from LHS into square root in RHS and determine the value of x first taking -5 and second time 5 in the expression.

Complete step by step solution:
We are given a mathematical expression having one variable xx in the term.
(x2)2=25{\left( {x - 2} \right)^2} = 25
Transposing the square from LHS to RHS it will become ±\pm \sqrt {}
x2=±25 x2=±5 x2=5 x=5+2 x=7 now x2=5 x=5+2 x=3  x - 2 = \pm \sqrt {25} \\\ x - 2 = \pm 5 \\\ x - 2 = 5 \\\ \Rightarrow x = 5 + 2 \\\ \Rightarrow x = 7 \\\ now \\\ x - 2 = - 5 \\\ \Rightarrow x = - 5 + 2 \\\ \Rightarrow x = - 3 \\\
The value of x can be 3,7 - 3,7
Therefore, solution to the expression (x2)2=25{\left( {x - 2} \right)^2} = 25is x=3,7x = - 3,7

Note Quadratic Equation: A quadratic equation is a equation which can be represented in the form of
ax2+bx+ca{x^2} + bx + cwhere xxis the unknown variable and a,b,c are the numbers known where a0a \ne 0.If a=0a = 0then the equation will become a linear equation and will no longer be quadratic .
The degree of the quadratic equation is of the order 2.
Discriminant: D=b24acD = {b^2} - 4ac
Using Discriminant, we can find out the nature of the roots
If D is equal to zero, then both of the roots will be the same and real.
If D is a positive number then, both of the roots are real solutions.
If D is a negative number, then the root are the pair of complex solutions
In order to determine the roots to a quadratic equation, there are couple of ways,

1.Using splitting up the middle term method:
let ax2+bx+ca{x^2} + bx + c
calculate the product of coefficient of x2{x^2}and the constant term and factorise it into two factors
in a way that either addition or subtraction of the two gives the middle term and multiplication gives the product value.
2.You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
x1=b+b24ac2ax1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} and x2=bb24ac2ax2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}
x1,x2 are root to quadratic equation ax2+bx+ca{x^2} + bx + c