<span>8p<84
p<84/8
p<10.5
answer
</span><span>8, 9, 10 are solutions</span>
The general form of a polynomial is: x² + bx + c = 0. Substituting the roots,
3² + b(3) + c = 0
3b + c = -9 --> eqn 1
(5 + √5)² + (5 +√5)(b) + c = 0
(5+√5)b + c = -30-10√5 --> eqn 2
Solving both equations simultaneously by subtracting eqn 2 from eqn 1,
(-2 - √5)b = 21 + 10√5
b = -8 - √5
Using eqn 1,
3(-8 - √5) + c = -9
-24 - 3√5 + c = -9
c = -9 + 24 + 3√5
c = 15 + 3√5
Hence, the polynomial is: <em>f(x) = x² + (-8 - √5)x + (15 + 3√5)</em>.
The answer is B.
When you set up a ratio of all the corresponding sides, the ratios all equal 2:1
Answer:
The answer is (d) ⇒ ![pq^{2}r\sqrt[3]{pr^{2}}](https://tex.z-dn.net/?f=pq%5E%7B2%7Dr%5Csqrt%5B3%5D%7Bpr%5E%7B2%7D%7D)
Step-by-step explanation:
* To simplify the cube roots:
If its number then the number must be written in the form x³
then we divide the power by 3 to cancel the radical
If its variable we divide its power by 3 to cancel the radical
∵ ![\sqrt[3]{p^{4}q^{6}r^{5}}=p^{\frac{4}{3}}q^{\frac{6}{3}}r^{\frac{5}{3}}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bp%5E%7B4%7Dq%5E%7B6%7Dr%5E%7B5%7D%7D%3Dp%5E%7B%5Cfrac%7B4%7D%7B3%7D%7Dq%5E%7B%5Cfrac%7B6%7D%7B3%7D%7Dr%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%7D)
∴ 
∵ ![p^{\frac{1}{3}}=\sqrt[3]{p}](https://tex.z-dn.net/?f=p%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3D%5Csqrt%5B3%5D%7Bp%7D)
∵ ![r^{\frac{2}{3}}=\sqrt[3]{r^{2}}](https://tex.z-dn.net/?f=r%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%3D%5Csqrt%5B3%5D%7Br%5E%7B2%7D%7D)
∴ ![p(p)^{\frac{1}{3}}q^{2}r(r)^{\frac{2}{3}}=p(\sqrt[3]{p})q^{2}r(\sqrt[3]{r^{2}})](https://tex.z-dn.net/?f=p%28p%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7Dq%5E%7B2%7Dr%28r%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%3Dp%28%5Csqrt%5B3%5D%7Bp%7D%29q%5E%7B2%7Dr%28%5Csqrt%5B3%5D%7Br%5E%7B2%7D%7D%29)
∴ ![prq^{2}\sqrt[3]{pr^{2}}}](https://tex.z-dn.net/?f=prq%5E%7B2%7D%5Csqrt%5B3%5D%7Bpr%5E%7B2%7D%7D%7D)
∴ The answer is (d)
