97+x=27+64
x=27+64-97
x=-6
(Hopefully this is correct)
Answer:
dum there is part A prat B and C
Step-by-step explanation:
<h2>
Hello!</h2>
The answers are:
The possible values for x in the equation, are:
First option, ![5\sqrt[3]{3}](https://tex.z-dn.net/?f=5%5Csqrt%5B3%5D%7B3%7D)
Second option, ![\sqrt[3]{375}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B375%7D)
<h2>
Why?</h2>
To solve the problem, we need to remember the following properties of the exponents and roots:
![a\sqrt[n]{b}=\sqrt[n]{a^{n}*b} \\\\\sqrt[n]{a^{m} }=a^{\frac{m}{n}}\\\\(a^{b})^{c}=a^{b*c}](https://tex.z-dn.net/?f=a%5Csqrt%5Bn%5D%7Bb%7D%3D%5Csqrt%5Bn%5D%7Ba%5E%7Bn%7D%2Ab%7D%20%5C%5C%5C%5C%5Csqrt%5Bn%5D%7Ba%5E%7Bm%7D%20%7D%3Da%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%5C%5C%5C%5C%28a%5E%7Bb%7D%29%5E%7Bc%7D%3Da%5E%7Bb%2Ac%7D)
Then, we are given the expression:
So, finding "x", we have:
![x^{3}=375\\\\(x^{3})^{\frac{1}{3} } =(375)^{\frac{1}{3}}\\\\x=\sqrt[3]{375}=\sqrt[3]{125*3}=\sqrt[3]{125}*\sqrt[3]{3}=5\sqrt[3]{3}](https://tex.z-dn.net/?f=x%5E%7B3%7D%3D375%5C%5C%5C%5C%28x%5E%7B3%7D%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D%20%3D%28375%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5C%5C%5C%5Cx%3D%5Csqrt%5B3%5D%7B375%7D%3D%5Csqrt%5B3%5D%7B125%2A3%7D%3D%5Csqrt%5B3%5D%7B125%7D%2A%5Csqrt%5B3%5D%7B3%7D%3D5%5Csqrt%5B3%5D%7B3%7D)
Hence, the possible values for x in the equation, are:
First option,
Second option,
Have a nice day!
Given :-
- a² - 2a - b² = 0
- 2b + 2ab = 0
To find :-
Solution :-
<u>Taking</u><u> </u><u>second</u><u> </u><u>equation</u><u>:</u><u>-</u>
- 2b + 2ab = 0
- 2b ( 1 + a ) = 0
- 2b = 0 or (1+a) = 0
- b = 0 , a = -1
<u>Substitute</u><u> </u><u>in </u><u>first </u><u>equation</u><u> </u><u>:</u><u>-</u><u> </u>
<u>When </u><u>b </u><u>=</u><u> </u><u>0</u><u> </u><u>,</u>
- a² - 2a - 0² = 0
- a² - a = 0
- a( a -1) =0
- a = 0 , 1
<u>When </u><u>a </u><u>=</u><u> </u><u>-</u><u>1</u><u> </u><u>,</u>
- (-1)² - 2*(-1) - b² = 0
- 1 + 2 - b² = 0
- b² = 3
- b = ±√3
<u>Answer </u><u>:</u><u>-</u><u> </u>
- a = 0,1 ; b = 0
- a = -1 , b = ±√3
