q 4 4q 3 8q 2 4q 1 0 Find the value of q

q^2(q+1)-4(q+1)=0

q^2-4)(q+1)=0

q-2)(q+2)(q+1)=0

then q-2=0 or q+2=0 or q+1=0

Untie the orange balance belt q=2 or trap -2 or -1

Decompose by cross multiplication to get (q-1)(2q+1)=0 to get q=1 or q=-1 2

3q 3-q 4=2q 4-3q 3+2=0q 4-q 3-2q 3+2q 2-2q 2+2q-2q+2=0(q-1)*q 3-2(q-1)*q 2-2(q-1)*q-2(q-1)*q-2(q-1)*(q 3-2q 2-2q-2)=0 and q≠1, so q 3-2q 2-2q-2=0 above one yuan cubic town Weicheng can pretend to be a brigade and then solve a real root about the hall or two conjugates.

Lou Xingzhi knows that the result is correct, and the process is a little bit of a problem: let p=q 3, then p+p 2=2p 3, factorization to get p(2p+1)(p-1)=0 (here should be :(2p+1)(p-1)=0), get p=0 or -1 2 or 1, so q=0 or 3 root number (-1 worse than 2) or 1

The original recipe is Zao Cheng to accompany the god for Sun to dismantle 2 q+2q=17

2+2q^2=17q

2q^2-17q+2=0

q=(17±√273)/4

Decompose the factor.

q^2(q+1)-4(q+1)=0

q^2-4）(q+1)=0

So there is q= 2 or q=-1

q^4-q^3-q^2=0

q^2(q^2-q-1)=0

q 2 = 0 or q 2 - q - 1 = 0

Q1=Q2=0,Q3=(1+Root5) 2,Q4=(1-Root5) 2Hope it helps.

If you have any questions, you can follow up.

Thank you.

The original formula can be turned into.

q^4-q^3-q^2=0

Q2(Q2-Q-1)=0

The solution gives q=0 or q 2-q-1=0

From q 2-q-1=0, we get (q-1 2) 2=5 4 q-1 2= 5 2 or q-1 2=- 5 2 q=(1+ 5) 2 or q=(1- 5) 2, so the value of q is: q1=0; q2=(1+√5)/2 ;q3=(1-√5)/2

q^2=q^2 *q^2-q^2 *q

q^2=q^2*(q^2-q)

1=q^2-q

The solution gives q=0 or q=(1 5)2

If you don't understand, you can ask. If you are satisfied

There are four solutions, q=(1 5)2 and two identical roots q=0