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for the first equation don't let the cube trip you up. simply factor out 5x because 5 goes into all three numbers evenly as does the x. so now your equation reads 5x(x²+6x+9)=0. now factor x²+6x+9 like you normally would. now you should have **3 possible roots. **5x=0, x+3=0, and x+3=0. once you solve for x you should have x=0 and x=-3.
for the second one its a little trickier. we cant factor out the way we did in number one so you try to get all the x's to one side. ⇒ x^4-4x²=-3. now you can factor x² out to get x²(x²-4)=-3. now solve for x!! x²=-3 and x²-4=-3. you get x=√-3, x=1 and x=-1
for the last one your going to solve the original version of the problem (2x-5=11) and the negated version of the problem. (-2x+5=11) all you're doing is solving for x. you should get x=-3 and x=8

Answered by secretlyblu | 2024-06-10

5 x 3 + 30 x 2 + 45 x = 0 x 3 + 6 x 2 + 9 x = 0 x ( x 2 + 6 x + 9 ) = 0 x ( x + 3 ) 2 = 0 x = 0 ∨ x = − 3 x 4 − 4 x 2 + 3 = 0 x 4 − x 2 − 3 x 2 + 3 = 0 x 2 ( x 2 − 1 ) − 3 ( x 2 − 1 ) = 0 ( x 2 − 3 ) ( x 2 − 1 ) = 0 ( x 2 − 3 ) ( x − 1 ) ( x + 1 ) = 0 x = − 3 ​ ∨ x = 3 ​ ∨ x = 1 ∨ x = − 1
∣2 x − 5∣ = 11 2 x − 5 = 11 ∨ 2 x − 5 = − 11 2 x = 16 ∨ 2 x = − 6 x = 8 ∨ x = − 3

Answered by konrad509 | 2024-06-10

We found solutions for three equations. The first equation has roots x = 0 and x = -3. The second equation has roots x = 1, -1, \sqrt{3}, and -\sqrt{3}, while the last equation has solutions x = 8 and x = -3.
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Answered by secretlyblu | 2025-01-29

Jawaban:- sering melakukan hal hal positiflebih dekat lagi dengan tuhan-sering berolahraga-bergaul dengan lingkungan yang baiksmgt!

Answered by aja657479 | 2025-07-02