Answer
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Hint: In this question, we need to factor the given polynomial. Note that the given polynomial is of degree 2, and hence it is a quadratic polynomial. It is of the form of $a{x^2} + bx + c$. We find the roots using the formula, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. Substitute for the values of a, b and c and obtain the required factors by simplifying and solving the given problem. Then obtain the desired result.
Complete step-by-step answer:
Given an equation of the form $5{x^2} - 4x - 2 = 0$
It is mentioned that we need to factor the given polynomial.
Note that the given equation is one of the examples for quadratic equations.
Consider an equation of the form $a{x^2} + bx + c$, where a, b, c are any real numbers.
We find the roots of this equation by using quadratic formula which is given by,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
In the given problem, we have $a = 5$, $b = - 4$ and $c = - 2$.
Hence substituting these values in the formula we get,
$ \Rightarrow x = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 5 \times ( - 2)} }}{{2 \times 5}}$
Simplifying this we get,
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {16 - 20 \times ( - 2)} }}{{10}}$
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {16 + 40} }}{{10}}$
$ \Rightarrow x = \dfrac{{5 \pm \sqrt {56} }}{{10}}$ …… (1)
We can write $\sqrt {56} = \sqrt {4 \times 14} $
This can be written as,
$ \Rightarrow \sqrt {56} = \sqrt 4 \times \sqrt {14} $
We know that $\sqrt 4 = 2$
Hence we get,
$ \Rightarrow \sqrt {56} = 2\sqrt {14} $
Thus the equation (1) becomes,
$ \Rightarrow x = \dfrac{{4 \pm 2\sqrt {14} }}{{10}}$
Now we rearrange and isolate the variable to find the solution.
We notice that the number 2 is common in both numerator and denominator.
Thus, taking out 2 in both numerator and denominator, we get,
$ \Rightarrow x = \dfrac{{2(2 \pm \sqrt {14} )}}{{2(5)}}$
Cancelling out 2 in both numerator and denominator we get,
$ \Rightarrow x = \dfrac{{2 \pm \sqrt {14} }}{5}$
Hence the solution for the equation $5{x^2} - 4x - 2 = 0$ is given by $x = \dfrac{{2 \pm \sqrt {14} }}{5}$.
Note:
Students may get confused when choosing the method to solve such a problem. The above problem cannot be solved using factoring methods. If the polynomial of the form $a{x^2} + bx + c$, then we can rewrite the middle term as a sum of terms whose product is $a \cdot c$ and whose sum is b.
Since in the given problem the middle term cannot be expressed as the sum of terms whose product is $a \cdot c$ and whose sum is b.
So we need to use the quadratic formula to find out the roots. The quadratic formula is given by,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Complete step-by-step answer:
Given an equation of the form $5{x^2} - 4x - 2 = 0$
It is mentioned that we need to factor the given polynomial.
Note that the given equation is one of the examples for quadratic equations.
Consider an equation of the form $a{x^2} + bx + c$, where a, b, c are any real numbers.
We find the roots of this equation by using quadratic formula which is given by,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
In the given problem, we have $a = 5$, $b = - 4$ and $c = - 2$.
Hence substituting these values in the formula we get,
$ \Rightarrow x = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 5 \times ( - 2)} }}{{2 \times 5}}$
Simplifying this we get,
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {16 - 20 \times ( - 2)} }}{{10}}$
$ \Rightarrow x = \dfrac{{4 \pm \sqrt {16 + 40} }}{{10}}$
$ \Rightarrow x = \dfrac{{5 \pm \sqrt {56} }}{{10}}$ …… (1)
We can write $\sqrt {56} = \sqrt {4 \times 14} $
This can be written as,
$ \Rightarrow \sqrt {56} = \sqrt 4 \times \sqrt {14} $
We know that $\sqrt 4 = 2$
Hence we get,
$ \Rightarrow \sqrt {56} = 2\sqrt {14} $
Thus the equation (1) becomes,
$ \Rightarrow x = \dfrac{{4 \pm 2\sqrt {14} }}{{10}}$
Now we rearrange and isolate the variable to find the solution.
We notice that the number 2 is common in both numerator and denominator.
Thus, taking out 2 in both numerator and denominator, we get,
$ \Rightarrow x = \dfrac{{2(2 \pm \sqrt {14} )}}{{2(5)}}$
Cancelling out 2 in both numerator and denominator we get,
$ \Rightarrow x = \dfrac{{2 \pm \sqrt {14} }}{5}$
Hence the solution for the equation $5{x^2} - 4x - 2 = 0$ is given by $x = \dfrac{{2 \pm \sqrt {14} }}{5}$.
Note:
Students may get confused when choosing the method to solve such a problem. The above problem cannot be solved using factoring methods. If the polynomial of the form $a{x^2} + bx + c$, then we can rewrite the middle term as a sum of terms whose product is $a \cdot c$ and whose sum is b.
Since in the given problem the middle term cannot be expressed as the sum of terms whose product is $a \cdot c$ and whose sum is b.
So we need to use the quadratic formula to find out the roots. The quadratic formula is given by,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
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