Question

Which quadratic equation equation is equivalent to (x^2-1)^2 -11(^2-1) +24=0

Answer 1

$(x^2-1)^2-11(x^2-1)+24=0$

Use substitution: $x^2-1=t$

$t^2-11t+24=0\\\\t^2-8t-3t+24=0\\\\t(t-8)-3(t-8)=0\\\\(t-8)(t-3)=0\iff t-8=0\ \vee\ t-3=0\\\\t=8\ \vee\ t=3$

we're going back to substitution:

$x^2-1=8\ \vee\ x^2-1=3\ \ \ \ |add\ 1\ to\ both\ sides\ of\ the\ equations\\\\x^2=9\ \vee\ x^2=4$

therefore

$x=\pm\sqrt9\ \vee\ x=\pm\sqrt4\\\\\boxed{x=-3\ \vee\ x=3\ \vee\ x=-2\ \vee\ x=2}$

Answer 2

Answer:

x = ± 3 and ± 2

Step-by-step explanation:

The given quadratic equation is

$(x^{2}-1)^{2}-11(x^{2}-1)+24=0$

To make this question easier we will consume ( x² -1 ) = a

a² - 11a + 24 = 0

Now we can factorize this equation easily

a²- 8a - 3a + 24 = 0

a (a-8) - 3 ( a-8) = 0

( a-3 ) ( a-8 ) = 0

Therefore, a - 3 = 0 ⇒ a = 3

or               a - 8 = 0 ⇒ a = 8

Now we put the value a

when a = 3    (x² - 1) = 3

                     x² = 3 + 1 = 4

                    x = √4

                       = ± 2

when a = 8   (x² - 1) = 8

                     x² = 9

                     x = √9

                        = ± 3

Therefore, x = ± 3 and ± 2 will be the answer.