[tex]x^4-13x^2+12=0\\[/tex]
Let [tex]x^2=x[/tex]
[tex]x^2-13x+12=0\\(x-12)(x-1)=0\\x=1,12\\x^2=1,12\\x=1,-1,\sqrt{12},-\sqrt{12}[/tex]
Answer:
x = ± 1, x = ± 2[tex]\sqrt{3}[/tex]
Step-by-step explanation:
Given
[tex]x^{4}[/tex] - 13x² + 12 = 0
Use the substitution u = x² , then
u² - 13u + 12 = 0 ← in standard form
(u - 1)(u - 12) = 0 ← in factored form
Equate each factor to zero and solve for u
u - 1 = 0 ⇒ u = 1
u - 12 = 0 ⇒ u = 12
Now substitute x² = u back, that is
x² = 1 ( take the square root of both sides )
x = ± [tex]\sqrt{1}[/tex] = ± 1
x² = 12 , then
x = ± [tex]\sqrt{12}[/tex] = ± [tex]\sqrt{4(3)}[/tex] = ± 2[tex]\sqrt{3}[/tex]
