Question
Question: How do you solve \({\left( {x - 2} \right)^2} = 25\)?...
How do you solve (x−2)2=25?
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 x in the term.
(x−2)2=25
Transposing the square from LHS to RHS it will become ±
x−2=±25 x−2=±5 x−2=5 ⇒x=5+2 ⇒x=7 now x−2=−5 ⇒x=−5+2 ⇒x=−3
The value of x can be −3,7
Therefore, solution to the expression (x−2)2=25is x=−3,7
Note Quadratic Equation: A quadratic equation is a equation which can be represented in the form of
ax2+bx+cwhere xis the unknown variable and a,b,c are the numbers known where a=0.If a=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=b2−4ac
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+c
calculate the product of coefficient of x2and 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=2a−b+b2−4ac and x2=2a−b−b2−4ac
x1,x2 are root to quadratic equation ax2+bx+c