Answer:
2 solutions
Step-by-step explanation:
x^4-6x^2-7=0
Replace x^2 by y
(x^2)^2-6x^2-7=0
y^2-6y-7=0
D=b^2-4ac= (-6)^2-4*1*(-7)= 36+28=64
sqrtD=8
y1=(-b-sqrtD)/2a=-2/2=-1
y2= (-b+sqrtD)/2a=14/2=7
x^2=-1(n0 roots) x^2=7 x=sqrt7 or x=-sqrt7
2 roots
Answer:
HL
Step-by-step explanation:
Use the HL Congruence Theorem to prove that the triangles are congruent.
You can first take out an x^2 from each number , 3x^6 - 39x^4 + 108x^2 = x^2 ( 3x^4 - 39x^2 + 108) Then just factor and solve for the zero's of the equation
