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Answer:
5. x = (6 +√7)/29; a=6, b=1, c=29
6. x = 2
Explanation:
5.
The quadratic formula can be used, where a=(5+4√7), b=(4-2√7), c=-1.
![x=(-b+√(b^2-4ac))/(2a)=\frac{-(4-2√(7))+\sqrt{(4-2√(7))^2-4(5+4√(7)})(-1)}{2(5+4√(7))}\\\\=\frac{-4+2√(7)+\sqrt{16-16√(7)+28+20+16√(7)}}{10+8√(7)}=(4+2√(7))/(2(5+4√(7)))\\\\=((2+√(7))(5-4√(7)))/((5+4√(7))(5-4√(7)))=(10-3√(7)-28)/(25-112)=\boxed{(6+√(7))/(29)}](https://img.qammunity.org/2022/formulas/mathematics/high-school/foutvnnn1lw8r4r9q31j9kz6y1f6u4j886.png)
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6.
Use the substitution z=3^x to put the equation in the form ...
z² -3z -54 = 0
(z -9)(z +6) = 0 . . . . . factor
z = 9 or -6 . . . . . . . . value of z that make the factors zero
Only the positive solution is useful, since 3^x cannot be negative.
z = 9 = 3^2 = 3^x . . . . use the value of z to find x
x = 2