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Sagot :
The roots of the given polynomials exist [tex]$x=8+\sqrt{10}$[/tex], and [tex]$x=8-\sqrt{10}$[/tex].
What is the formula of the quadratic equation?
For a quadratic equation of the form [tex]$a x^{2}+b x+c=0$[/tex] the solutions are
[tex]$x_{1,2}=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$[/tex]
Therefore by using the formula we have
[tex]$x^{2}-16 x+54=0$[/tex]
Let, a = 1, b = -16 and c = 54
Substitute the values in the above equation, and we get
[tex]$x_{1,2}=\frac{-(-16) \pm \sqrt{(-16)^{2}-4 \cdot 1 \cdot 54}}{2 \cdot 1}$[/tex]
simplifying the equation, we get
[tex]${data-answer}amp;x_{1,2}=\frac{-(-16) \pm 2 \sqrt{10}}{2 \cdot 1} \\[/tex]
[tex]${data-answer}amp;x_{1}=\frac{-(-16)+2 \sqrt{10}}{2 \cdot 1}, x_{2}=\frac{-(-16)-2 \sqrt{10}}{2 \cdot 1} \\[/tex]
[tex]${data-answer}amp;x=8+\sqrt{10}, x=8-\sqrt{10}[/tex]
Therefore, the roots of the given polynomials are [tex]$x=8+\sqrt{10}$[/tex], and
[tex]$x=8-\sqrt{10}$[/tex].
To learn more about quadratic equations refer to:
brainly.com/question/1214333
#SPJ4
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