Answer: 6(x²-4x+4-4)+1=0, 6(x-2)²-24+1=0, 6(x-2)²=23, x-2=±√(23/6), x=2±√(23/6)=2±1.95789, so x=3.95789 or 0.04211 approx. these are the zeros.
step by step explanation:
\boxed{\boxed{\dfrac{12+\sqrt{138}}{6},\ \dfrac{12-\sqrt{138}}{6}}}
Solution-
The quadratic function is,
6x^2-24x + 1
a = 6, b = -24, c = 1
x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}
=\dfrac{-(-24)\pm \sqrt{-24^2-4\cdot 6\cdot 1}}{2\cdot 6}
=\dfrac{24\pm \sqrt{576-24}}{12}
=\dfrac{24\pm \sqrt{552}}{12}
=\dfrac{24\pm 2\sqrt{138}}{12}
=\dfrac{12\pm \sqrt{138}}{6}
=\dfrac{12+\sqrt{138}}{6},\ \dfrac{12-\sqrt{138}}{6}
The correct answer is 117
x + y = 236
y = x + 2
x + x + 2 = 236
2x = 236 - 2
2x = 234
x = 234/2
x = 117
y = 117 + 2
y =119
Both are odd integers and 117 is smaller.
For this case we have the following expression:
For power properties we have:
Rewriting the exponents of the expression we have:
Using the cubic root we have:
![(\frac{1}{\sqrt[3]{8^2}} a^2)](https://tex.z-dn.net/?f=%20%28%5Cfrac%7B1%7D%7B%5Csqrt%5B3%5D%7B8%5E2%7D%7D%20a%5E2%29%20%20%20)
![(\frac{1}{\sqrt[3]{64}} a^2)](https://tex.z-dn.net/?f=%20%28%5Cfrac%7B1%7D%7B%5Csqrt%5B3%5D%7B64%7D%7D%20a%5E2%29%20%20%20)
![(\frac{1}{\sqrt[3]{4^3}} a^2)](https://tex.z-dn.net/?f=%20%28%5Cfrac%7B1%7D%7B%5Csqrt%5B3%5D%7B4%5E3%7D%7D%20a%5E2%29%20%20%20)
Simplifying the expression we have:
Answer:
The equivalent expression is given by:
For the first 1 it is d and the second one is d the third one is d
