Question

What substitution should be used to rewrite 16(x3 + 1)2 – 22(x3 + 1) – 3 = 0 as a quadratic equation?

Answer 1

Answer:

u = (x3 + 1)

Step-by-step explanation:

This is correct, I just took the test! I hope this helps!

(this is the only place I could find this question, and this is the only correct answer to it)

Answer 2

ANSWER:  

We need the substitution of $x^{3}$ + 1 = t, to make given equation into quadratic equation.

SOLUTION:

Given, equation is  $16\left(x^{3}+1\right)^{2}-22\left(x^{3}+1\right)-3=0$ → (1)

Above given equation is having highest power as 6, so it is not an quadratic equation.

We need to perform an substitution so that above equation turns to quadratic equation.

We know that, general form of a quadratic equation is given by :

$a x^{2}+b x+c=0, \text { where } a \neq 0 \text { and } a, b, c \text { are constants }$

Now, compare the general form and the equation we have.

$\text { When }\left(x^{3}+1\right) \text { is considered as a single unit, equation will have degree } 2$

$\text { So, let us put } x^{3}+1=t \text { in }(1)$

$\text { Then, }(1) \rightarrow 16 t^{2}-22 t-3=0$

Where, a = 16, b = -22 and c =-3

Hence, we need the substitution of $x^{3}$ + 1 = t, to make given equation into quadratic equation.