
How do you solve \[{x^2} - 2x - 2\] ?
Answer
544.8k+ views
Hint: Here in this question, we have to solve the given quadratic equation and find the roots for the variable x. we can solve above equation by factoring or by the considering or applying the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] or we can also use sum product rule. Hence, we obtain the required result.
Complete step-by-step answer:
The equation is in the form of a quadratic equation. The degree of the equation is 2. Therefore, we obtain two values after simplifying. Here we solve the equation by using formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
In general, the quadratic equation will be in the form of \[a{x^2} + bx + c\] , where a, b, c are constants and the x is variable.
Now consider the equation \[{x^2} - 2x - 2\]
When we compare the given equation to the general form of equation we have a = 1, b = -2 and c = -2.
Substituting these values in the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] we get
\[ \Rightarrow x = \dfrac{{ - ( - 2) \pm \sqrt {{{( - 2)}^2} - 4(1)( - 2)} }}{{2(1)}}\]
On simplifying we get
\[ \Rightarrow x = \dfrac{{2 \pm \sqrt {4 + 8} }}{2}\]
On further simplification we get
\[ \Rightarrow x = \dfrac{{2 \pm \sqrt {12} }}{2}\]
The number 12 is not a perfect square. So the square root 12 can be written as
\[ \Rightarrow x = \dfrac{{2 \pm \sqrt {4 \times 3} }}{2}\]
The number is a perfect square so let we take 4 outside from the square root. So we have
\[ \Rightarrow x = \dfrac{{2 \pm 2\sqrt 3 }}{2}\]
The two terms in the numerator are having 2. So we can take common, and it is written as
\[ \Rightarrow x = \dfrac{{2(1 \pm \sqrt 3 )}}{2}\]
We cancel the number 2, which is present in the both numerator and in the denominator.
\[ \Rightarrow x = 1 \pm \sqrt 3 \]
Therefore we have
\[ \Rightarrow x = 1 + \sqrt 3 \] and \[x = 1 - \sqrt 3 \]
Hence we have solved the given equation and found the roots for the equation.
So, the correct answer is “ \[ \Rightarrow x = 1 + \sqrt 3 \] and \[x = 1 - \sqrt 3 \] ”.
Note: The equation is a quadratic equation. This problem can be solved by using the sum product rule. This defines as for the general quadratic equation \[a{x^2} + bx + c\] , the product of \[a{x^2}\] and c is equal to the sum of bx of the equation or we can solve the equation by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] . Sometimes the factorisation method is not applicable to solve the quadratic equation.
Complete step-by-step answer:
The equation is in the form of a quadratic equation. The degree of the equation is 2. Therefore, we obtain two values after simplifying. Here we solve the equation by using formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
In general, the quadratic equation will be in the form of \[a{x^2} + bx + c\] , where a, b, c are constants and the x is variable.
Now consider the equation \[{x^2} - 2x - 2\]
When we compare the given equation to the general form of equation we have a = 1, b = -2 and c = -2.
Substituting these values in the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] we get
\[ \Rightarrow x = \dfrac{{ - ( - 2) \pm \sqrt {{{( - 2)}^2} - 4(1)( - 2)} }}{{2(1)}}\]
On simplifying we get
\[ \Rightarrow x = \dfrac{{2 \pm \sqrt {4 + 8} }}{2}\]
On further simplification we get
\[ \Rightarrow x = \dfrac{{2 \pm \sqrt {12} }}{2}\]
The number 12 is not a perfect square. So the square root 12 can be written as
\[ \Rightarrow x = \dfrac{{2 \pm \sqrt {4 \times 3} }}{2}\]
The number is a perfect square so let we take 4 outside from the square root. So we have
\[ \Rightarrow x = \dfrac{{2 \pm 2\sqrt 3 }}{2}\]
The two terms in the numerator are having 2. So we can take common, and it is written as
\[ \Rightarrow x = \dfrac{{2(1 \pm \sqrt 3 )}}{2}\]
We cancel the number 2, which is present in the both numerator and in the denominator.
\[ \Rightarrow x = 1 \pm \sqrt 3 \]
Therefore we have
\[ \Rightarrow x = 1 + \sqrt 3 \] and \[x = 1 - \sqrt 3 \]
Hence we have solved the given equation and found the roots for the equation.
So, the correct answer is “ \[ \Rightarrow x = 1 + \sqrt 3 \] and \[x = 1 - \sqrt 3 \] ”.
Note: The equation is a quadratic equation. This problem can be solved by using the sum product rule. This defines as for the general quadratic equation \[a{x^2} + bx + c\] , the product of \[a{x^2}\] and c is equal to the sum of bx of the equation or we can solve the equation by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] . Sometimes the factorisation method is not applicable to solve the quadratic equation.
Recently Updated Pages
Master Class 9 Social Science: Engaging Questions & Answers for Success

Master Class 9 Science: Engaging Questions & Answers for Success

Master Class 9 English: Engaging Questions & Answers for Success

Master Class 9 Maths: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Class 9 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
Which places in India experience sunrise first and class 9 social science CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Write the 6 fundamental rights of India and explain in detail

Difference Between Plant Cell and Animal Cell

What is pollution? How many types of pollution? Define it

What is the Full Form of ISI and RAW

