How do you find the zeros of a function algebraically, $ f\left( x \right) = {x^2} - 14x - 4 $ ?
Answer
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Hint: In this question we need to determine the zeros of the function $ f\left( x \right) = {x^2} - 14x - 4 $ . Here, first we will understand what the zero of a function means. Then, we will use the quadratic formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ . We will compare the given function with the standard form of the quadratic equation, by which we will find the values of $ a,\,b,\,c $ . Then, we will substitute the value in the formula by evaluating it to determine the roots of the given function.
Complete step by step answer:
Here, we need to determine the zeros of a function algebraically, $ f\left( x \right) = {x^2} - 14x - 4 $ .
First, let us know what is known as the zero of the function.
The zero of the function is defined as the point in which the value of the function is zero i.e., a root of a polynomial is a zero of the corresponding polynomial function.
Now, let us use the quadratic formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ where the standard form quadratic is $ a{x^2} + bx + c = 0 $ .
Therefore, comparing the given function with the standard form of the quadratic equation, we have,
$ a = 1 $
$ b = - 14 $
And, $ c = - 4 $
Now, substituting these values in the formula, we have,
$\Rightarrow$ $ x = \dfrac{{ - \left( { - 14} \right) \pm \sqrt {{{\left( { - 14} \right)}^2} - 4 \times 1 \times - 4} }}{{2 \times 1}} $
$\Rightarrow$ $ x = \dfrac{{14 \pm \sqrt {196 + 16} }}{2} $
$\Rightarrow$ $ x = \dfrac{{14 \pm \sqrt {212} }}{2} $
$\Rightarrow$ $ x = \dfrac{{14 \pm \sqrt {4 \times 53} }}{2} $
$\Rightarrow$ $ x = \dfrac{{14 \pm 2\sqrt {53} }}{2} $
$\Rightarrow$ $ x = \dfrac{{2\left( {7 \pm \sqrt {53} } \right)}}{2} $
$\Rightarrow$ Therefore, $ x = 7 \pm \sqrt {53} $
Hence, the zeros of the function $ f\left( x \right) = {x^2} - 14x - 4 $ is $ 7 + \sqrt {53} $ and $ 7 - \sqrt {53} $ .
Note:
In this question it is important to note here that a quadratic equation is an equation in the form of $ a{x^2} + bx + c = 0 $ where $ a $ is not equal to zero. The ‘roots’ of the quadratic are the numbers that satisfy the quadratic equation. There are always two roots for any quadratic equation, although sometimes they may coincide. And, the ‘solutions’ to the quadratic equation are where it is equal to zero. They are also called ‘rots’, or sometimes ‘zeros’.
Complete step by step answer:
Here, we need to determine the zeros of a function algebraically, $ f\left( x \right) = {x^2} - 14x - 4 $ .
First, let us know what is known as the zero of the function.
The zero of the function is defined as the point in which the value of the function is zero i.e., a root of a polynomial is a zero of the corresponding polynomial function.
Now, let us use the quadratic formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ where the standard form quadratic is $ a{x^2} + bx + c = 0 $ .
Therefore, comparing the given function with the standard form of the quadratic equation, we have,
$ a = 1 $
$ b = - 14 $
And, $ c = - 4 $
Now, substituting these values in the formula, we have,
$\Rightarrow$ $ x = \dfrac{{ - \left( { - 14} \right) \pm \sqrt {{{\left( { - 14} \right)}^2} - 4 \times 1 \times - 4} }}{{2 \times 1}} $
$\Rightarrow$ $ x = \dfrac{{14 \pm \sqrt {196 + 16} }}{2} $
$\Rightarrow$ $ x = \dfrac{{14 \pm \sqrt {212} }}{2} $
$\Rightarrow$ $ x = \dfrac{{14 \pm \sqrt {4 \times 53} }}{2} $
$\Rightarrow$ $ x = \dfrac{{14 \pm 2\sqrt {53} }}{2} $
$\Rightarrow$ $ x = \dfrac{{2\left( {7 \pm \sqrt {53} } \right)}}{2} $
$\Rightarrow$ Therefore, $ x = 7 \pm \sqrt {53} $
Hence, the zeros of the function $ f\left( x \right) = {x^2} - 14x - 4 $ is $ 7 + \sqrt {53} $ and $ 7 - \sqrt {53} $ .
Note:
In this question it is important to note here that a quadratic equation is an equation in the form of $ a{x^2} + bx + c = 0 $ where $ a $ is not equal to zero. The ‘roots’ of the quadratic are the numbers that satisfy the quadratic equation. There are always two roots for any quadratic equation, although sometimes they may coincide. And, the ‘solutions’ to the quadratic equation are where it is equal to zero. They are also called ‘rots’, or sometimes ‘zeros’.
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