
How do you factor $ {x^2} - x - 54 $ ?
Answer
533.4k+ views
Hint: Here we use the standard quadratic equation and will find the roots of the equation comparing the given equation with the standard quadratic equations- $ a{x^2} + bx + c = 0 $ where roots will be defined as \[x = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}\]
Complete step by step solution:
Take the given equation –
$ {x^2} - x - 54 $
Arrange in the form of the standard equation –
$ {x^2} - x - 54 = 0 $
Compare the above equation with the standard equation: $ a{x^2} + bx + c = 0 $
$
\Rightarrow a = 1 \\
\Rightarrow b = - 1 \\
\Rightarrow c = - 54 \;
$
Also, $ \Delta = {b^2} - 4ac $
Place the values from the given comparison
$ \Rightarrow \Delta = {( - 1)^2} - 4(1)( - 54) $
Simplify the above equation –
$ \Rightarrow \Delta = 1 + 216 $
Do Addition since negative and negative is positive. –
$ \Rightarrow \Delta = 217 $
Take square root on both the sides of the equation –
\[ \Rightarrow \sqrt \Delta = \sqrt {217} \]
The above expression can not be further reduced since the number on the right is not the perfect square.
Now, roots of the given equation can be expressed as –
\[x = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}\]
Place values in the above equation –
\[x = \dfrac{{ - ( - 1) \pm \sqrt {217} }}{2}\]
Product of minus and minus is plus. Simplify the above equation –
\[x = \dfrac{{1 \pm \sqrt {217} }}{2}\]
Therefore \[x = \dfrac{{1 - \sqrt {217} }}{2}\] or \[x = \dfrac{{1 - \sqrt {217} }}{2}\]
This is the required solution.
So, the correct answer is “ \[x = \dfrac{{1 - \sqrt {217} }}{2}\] or \[x = \dfrac{{1 - \sqrt {217} }}{2}\]”.
Note: Every quadratic polynomial has almost the two roots. Remember, the quadratic equations having coefficients as the rational numbers has the irrational roots. Also, the quadratic equations whose coefficients are all the distinct irrationals but both the roots are the rational.
Complete step by step solution:
Take the given equation –
$ {x^2} - x - 54 $
Arrange in the form of the standard equation –
$ {x^2} - x - 54 = 0 $
Compare the above equation with the standard equation: $ a{x^2} + bx + c = 0 $
$
\Rightarrow a = 1 \\
\Rightarrow b = - 1 \\
\Rightarrow c = - 54 \;
$
Also, $ \Delta = {b^2} - 4ac $
Place the values from the given comparison
$ \Rightarrow \Delta = {( - 1)^2} - 4(1)( - 54) $
Simplify the above equation –
$ \Rightarrow \Delta = 1 + 216 $
Do Addition since negative and negative is positive. –
$ \Rightarrow \Delta = 217 $
Take square root on both the sides of the equation –
\[ \Rightarrow \sqrt \Delta = \sqrt {217} \]
The above expression can not be further reduced since the number on the right is not the perfect square.
Now, roots of the given equation can be expressed as –
\[x = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}\]
Place values in the above equation –
\[x = \dfrac{{ - ( - 1) \pm \sqrt {217} }}{2}\]
Product of minus and minus is plus. Simplify the above equation –
\[x = \dfrac{{1 \pm \sqrt {217} }}{2}\]
Therefore \[x = \dfrac{{1 - \sqrt {217} }}{2}\] or \[x = \dfrac{{1 - \sqrt {217} }}{2}\]
This is the required solution.
So, the correct answer is “ \[x = \dfrac{{1 - \sqrt {217} }}{2}\] or \[x = \dfrac{{1 - \sqrt {217} }}{2}\]”.
Note: Every quadratic polynomial has almost the two roots. Remember, the quadratic equations having coefficients as the rational numbers has the irrational roots. Also, the quadratic equations whose coefficients are all the distinct irrationals but both the roots are the rational.
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