
How do you solve ${\left( {x - 2} \right)^2} = 25$?
Answer
557.7k+ views
Hint: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.
${\left( {x - 2} \right)^2} = 25$
Transposing the square from LHS to RHS it will become $ \pm \sqrt {} $
$
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$
Therefore, solution to the expression ${\left( {x - 2} \right)^2} = 25$is $x = - 3,7$
NoteQuadratic Equation: A quadratic equation is a equation which can be represented in the form of
$a{x^2} + bx + c$where $x$is the unknown variable and a,b,c are the numbers known where $a \ne
0$.If $a = 0$then 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: $D = {b^2} - 4ac$
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 $a{x^2} + bx + c$
calculate the product of coefficient of ${x^2}$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
$x1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
x1,x2 are root to quadratic equation $a{x^2} + bx + c$
Complete step by step solution:
We are given a mathematical expression having one variable $x$ in the term.
${\left( {x - 2} \right)^2} = 25$
Transposing the square from LHS to RHS it will become $ \pm \sqrt {} $
$
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$
Therefore, solution to the expression ${\left( {x - 2} \right)^2} = 25$is $x = - 3,7$
NoteQuadratic Equation: A quadratic equation is a equation which can be represented in the form of
$a{x^2} + bx + c$where $x$is the unknown variable and a,b,c are the numbers known where $a \ne
0$.If $a = 0$then 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: $D = {b^2} - 4ac$
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 $a{x^2} + bx + c$
calculate the product of coefficient of ${x^2}$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
$x1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
x1,x2 are root to quadratic equation $a{x^2} + bx + c$
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