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How do you solve \[{(x - 3)^2} = 25\]?

seo-qna
Last updated date: 29th Feb 2024
Total views: 340.8k
Views today: 9.40k
IVSAT 2024
Answer
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Hint: Here in this question we have to simplify the given algebraic equation. To simplify the above algebraic equation we use arithmetic operation multiplication. It is also simplified by using the standard algebraic formula \[{(a - b)^2} = {a^2} - 2ab + {b^2}\]. Hence we can obtain the solution for the question.

Complete step-by-step solution:
The algebraic expression or equation is a combination of variables, constants and arithmetic operations.
The equation can be simplified by using two methods.
Method 1:
Now consider the given equation \[{(x - 3)^2} = 25\]
The exponential form can be expanded
\[ \Rightarrow (x - 3)(x - 3) = 25\]
The terms in the braces are multiplied. On multiplying we get
\[ \Rightarrow x(x - 3) - 3(x - 3) = 25\]
On term-by-term multiplication
\[ \Rightarrow {x^2} - 3x - 3x + 9 = 25\]
On simplifying we get
\[ \Rightarrow {x^2} - 6x - 16 = 0\]
We can solve for x by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
So we have
\[
   \Rightarrow x = \dfrac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4(1)( - 16)} }}{{2(1)}} \\
   \Rightarrow x = \dfrac{{6 \pm \sqrt {36 + 64} }}{2} \\
   \Rightarrow x = \dfrac{{6 \pm \sqrt {100} }}{2} \\
   \Rightarrow x = \dfrac{{6 \pm 10}}{2} \\
 \]
So we have
\[ \Rightarrow x = \dfrac{{6 + 10}}{2}\] and \[x = \dfrac{{6 - 10}}{2}\]
Therefore
\[ \Rightarrow x = 8\] and \[x = - 2\]
Hence we have simplified the given equation
Method 2:
Now consider the given algebraic equation \[{(x - 3)^2} = 25\]
Apply the square root on the both sides of the equation we get
\[ \Rightarrow \sqrt {{{(x - 3)}^2}} = \sqrt {25} \]
In the LHS of the above equation the square and square root will get cancels and in the rhs the number 25 is a perfect square so we have
\[ \Rightarrow (x - 3) = \pm 5\]
Suppose in the RHS we have +5
\[ \Rightarrow x - 3 = 5\]
Taking -3 to RHS and simplifying we get
\[ \Rightarrow x = 5 + 3 = 8\]
Suppose in the RHS we have -5
\[ \Rightarrow x - 3 = - 5\]
Taking -3 to RHS and simplifying we get
\[ \Rightarrow x = - 5 + 3 = - 2\]
Hence we have simplified the given equation

Solving the equation by method 1 and method 2 we have obtained the final answer as the same.

Note: To multiply we use operation multiplication, multiplication of numbers is different from the multiplication of algebraic expression. In the algebraic expression it involves the both number that is constant and variables. Variables are also multiplied, if the variable is the same then the result will be in the form of exponent.
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