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Sagot :
Answer:
c) k = - 2
Step-by-step explanation:
Equação de Segundo Grau:
Uma equação de segundo grau tem o seguinte formato:
[tex]ax^2 + bx + c = 0[/tex]
Em que [tex]a \neq 0[/tex]
Caso tenham potências de x maiores que dois, o termo que multiplica essas potências deve ser 0.
Neste exercício:
Tem-se a seguinte equação:
[tex](k^2 - 4)x^3 + (k-2)x^2 + 7x - 8 = 0[/tex]
Para ser do segundo grau, necessita-se que:
[tex]k^2 - 4 = 0[/tex]
[tex]k^2 = 4[/tex]
[tex]k = \pm \sqrt{4}[/tex]
[tex]k = \pm 2[/tex]
E também
[tex]k - 2 \neq 0[/tex]
[tex]k \neq 2[/tex]
A intersecção entre [tex]k = \pm 2[/tex] e [tex]k \neq 2[/tex] é [tex]k = -2[/tex], e portanto a resposta é dada pela alternativa c.
O valor de k que torna a equação uma equação de segundo grau para a equação considerada é dado por: Opção C: k = -2
What is degree of an equation?
For a single variable equation, degree of an equation is the highest power the variable possess, such that its coefficient is non-zero.
A second degree equation is that equation where the maximum power involved in that equation is 2 (such that its coefficient is non-zero).
For this case, the considered equation is:
(k² – 4) x³ + ( k – 2 )x² + 7x - 8 = 0
For making it second degree, we would have to vanish the coefficient of x³, and by vanish, we mean we would like to make it 0, but simultaneously, we'd have to take care that coefficient of x² doesn't become 0 as we need it to be non-zero.
So we get two mathematical statements:
- Coefficient of x³ = 0
- Coefficient of x² ≠ 0
or
[tex](k^2 - 4) = 0\\k - 2 \neq 0[/tex]
Taking first equation, we get:
[tex]k^2 -4 = 0\\k^2 = 4\\k = \pm \sqrt{4} = \pm 2[/tex]
Thus, k = 2 or k = -2
From second statement, we get:
[tex]k -2 \neq 0\\\text{adding 2 on both the sides}\\k \neq 2[/tex]
Thus, from both the results, we conclude that k = -2 is correct choice.
Thus, the value of k that makes the equation a second degree equation for the considered equation is given by: Option C: k = -2
Learn more about second degree equation here:
https://brainly.com/question/17167165
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