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Yuri thinks 3/4 is a root of the following function.
q(x)=6x^3+19x^2-15x-28
Explain to Yuri why 3/4 cannot be a root.

Sagot :

Answer:

Since [tex]q(\frac{3}{4}) \neq 0[/tex], 3/4 cannot be a root of q(x).

Step-by-step explanation:

Root of a function:

If [tex]x^{\ast}[/tex] is a root of a function f(x), we have that [tex]f(x^{\ast}) = 0[/tex].

In this question.

We have to find [tex]q(\frac{3}{4})[/tex]. So

[tex]q(\frac{3}{4}) = 6(\frac{3}{4})^3 + 19(\frac{3}{4})^2 - 15(\frac{3}{4}) - 28[/tex]

[tex]q(\frac{3}{4}) = \frac{162}{64} + \frac{171}{16} - \frac{45}{4} - 28[/tex]

[tex]q(\frac{3}{4}) = \frac{162}{64} + \frac{684}{64} - \frac{720}{64} - \frac{1792}{64}[/tex]

[tex]q(\frac{3}{4}) \neq 0[/tex]

Since [tex]q(\frac{3}{4}) \neq 0[/tex], 3/4 cannot be a root of q(x).

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