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Write 6(x – 5)4 + 4(x – 5)2 + 6 = 0 in the form of a quadratic by using substitution.

A. 6u2 + 4u + 6 = 0, where u = x – 5
B. 6u2 + 4u + 6 = 0, where u = (x – 5)2
C. 6u4 + 4u + 6 = 0, where u = x – 5
D.6u4 + 4u + 6 = 0, where u = (x – 5)2

The answer is B. 6u2 + 4u + 6 = 0, where u = (x – 5)2 

Respuesta :

We are given the equation:

[tex]6(x-5)^{4}+4(x-5)^{2}+6=0[/tex]..........(1)

Now we have to write it in quadratic form using substitution.

The general form of quadratic equation is given by:

[tex]ax^{2}+bx+c=0[/tex]

So let us say

[tex](x-5)^{2}=u[/tex].......(2)

Plugging the value of (x-5)² from equation (2) in (1),

[tex]6u^{2}+4u+6=0[/tex]

Answer : Option B. [tex]6u^{2}+4u+6=0[/tex]


saryul

Answer:

B

Step-by-step explanation:

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