Answer:
9pr = 2q^2
Step-by-step explanation:
28.
px^2 + qx + r = 0
If one root is A then the other root is 2A
We have A + 2A = -q/p and A*2A = r/p
So 3A = -q/p so A = -q/3p and A^2 = r/2p so A = √ (r / 2p)
Equating these 2 equations:-
-q/3p = √(r / 2p)
q^2 / 9p^2 = r / 2p
9p^2r = 2pq^2
9pr = 2q^2 (answer)
I believe the first and last one are polynomials. Usually they start with the highest exponent first and then work their way down.
Answer:
Step-by-step explanation:
If you call "5x-2x^2+1" an "equation," then you must equate 5x-2x^2+1 to 0:
5x-2x^2+1 = 0
This is a quadratic equation. Rearranging the terms in descending order by powers of x, we get:
-2x^2 + 5x + 1 = 0. Here the coefficients are a = -2, b = 5 and c = 1.
Use the quadratic formula to solve for x:
First find the discriminant, b^2 - 4ac: 25 - 4(-2)(1) = 25 + 8 = 33
Because the discriminant is positive, the roots of this quadratic are real and unequal.
-b ± √(discriminant)
Applying the quadratic formula x = --------------------------------
2a
we get:
-5 ± √33 -5 + √33
x = ----------------- = --------------------- and
2(-2) -4
-5 - √33
---------------
-4
