pls help me solve for the roots of the quadratic trinomials

2. This isn't a homework service. You have to show that you at least made an effort to try them rather than just getting other people to do it for you.

We need proof that you actually tried, and then we can help you on parts that you get stuck on or cannot do.

3. HINT:
ax^4 + bx^2 + c = 0
Let k = x^2; then:
ak^2 + bk + c = 0

4. Those are NOT "quadratic trinomials" but they are of "quadratic type". That is, they can be changed to quadratic trinomials by a substitution like the one Wilmer used. Try it.

5. i can do for number 1 only

x^4 - 11x^2 + 28 = 0
(x^2 -7 ) ( x^2 -4 ) =
(x^2 - 7 ) 0 | (x^2 - 4) = 0
x^2 = 7 | x^2 = 4
x = = + - sq. root of 7| x = +- of 2

i dnt knw the rest beacuse i only knw the simple special products and factoring patterns.

2. x^-4 - 13x^-2 + 36 = 0
( x^-2 + 6 )(x^-2 + 6 ) =
i dnt knw next because it has a negative exponents and some of the rest of the given... sorry

rotchkaroots lang.

6. Originally Posted by rcs
2. x^-4 - 13x^-2 + 36 = 0
Check that equation: I'm sure it should be x^-4 - 12x^-2 + 36 = 0

Please use complete words; "dnt knw" are not words.

7. Originally Posted by rcs
i can do for number 1 only

x^4 - 11x^2 + 28 = 0
(x^2 -7 ) ( x^2 -4 ) =
(x^2 - 7 ) 0 | (x^2 - 4) = 0
x^2 = 7 | x^2 = 4
x = = + - sq. root of 7| x = +- of 2

i dnt knw the rest beacuse i only knw the simple special products and factoring patterns.

2. x^-4 - 13x^-2 + 36 = 0
( x^-2 + 6 )(x^-2 + 6 ) =
i dnt knw next because it has a negative exponents and some of the rest of the given... sorry

rotchkaroots lang.
Go back to x^-4- 13x^-2+ 36= 0.

There are two ways to handle that:
1) Let y= x^-2. Then x^-4= y^2 so the equation becomes y^2- 13y+ 36= 0. Once you have found y, x= 1/sqrt(y).

2) Multiply the entire equation by x^4. You get 1- 13x^2+ 36x^4= 0 which becomes a quadratic by letting y= x^2: 36y^2- 13y+ 1= 0. Once you have found y, x= sqrt(y).

8. how is number 3 and 4 done sir... i tried to make the fraction with an exponent of 2... i really can hardly do it... how will i able to make the first term be with an exponent of 2 and make 1 for the second variable term.

thnaks

9. Originally Posted by rcs
how is number 3 and 4 done sir... i tried to make the fraction with an exponent of 2... i really can hardly do it... how will i able to make the first term be with an exponent of 2 and make 1 for the second variable term.

thnaks
$\displaystyle \displaystyle x^{2/3}=\left(x^{1/3}\right)^2$

10. im sorry if i have made u think that way (Educated) i just needed help. i never needed an insult sir/madam. i also tried my best to solve the problems in my own little way. I thank you so much for your help. it just happend that i wasn't able to show to you how i did it. Sorry.

11. Originally Posted by rcs
im sorry if i have made u think that way (Educated) i just needed help. i never needed an insult sir/madam. i also tried my best to solve the problems in my own little way. I thank you so much for your help. it just happend that i wasn't able to show to you how i did it. Sorry.
"Educated" DID NOT insult you...he simply stated the rules that apply here.

12. ## pls check

am i doing it right... pls check my work pls

13. ## pls check

pls check my work

14. You can check them yourself just by putting the numbers into the original equation. If x= -23/9, then x+ 3= -23/9+ 27/9= 4/9 and $\displaystyle \sqrt{x+ 3}= 2/3$.

$\displaystyle 3(x+ 3)+ \sqrt{x+ 3}= 3(4/9) + 2/3= 4/3+ 2/3= 6/3= 2$ so that is correct.

If x= -2, then x+ 3= 1 and $\displaystyle \sqrt{x+3}= 1$.

$\displaystyle 3(x+ 3)+ \sqrt{x+ 3}= 3(1)+ 1= 4$, not 2, so that is not correct.

The reason you got an incorrect answer was that, after getting M= 2/3 and -1 as the (correct) solutions to the quadratic, you set $\displaystyle \sqrt{x+ 3}= -1$ which is impossible- a square root is never negative.