Question

Hey can you please help me posted picture of question :)

Answer 1

We can use quadratic formula to solve this equation.

The quadratic formula is:


`x= \frac{-b+- \sqrt{ b^(2)-4ac } }{2a}`

b = coefficient of x-term = -1
a= coefficient of squared term = 4
c = constant = 5

Using the values in the formula, we get:


`x= \frac{-(-1)+- \sqrt{ (-1)^(2)-4(4)(5) } }{2(4)} \\ \\ x= (1+- √((-79)) )/(8)`

Thus the two x-values that are solutions to the given equation are shown in option C and E

Answer 2

4x^2-x+5=0
We can use the quadratic formula
ax^2+bx+c=0; a=4, b=-1, c=5
x=[-b+-sqrt(b^2-4ac)]/(2a)
x=[-(-1)+-sqrt((-1)^2-4(4)(5)]/[2(4)]
x=[1+-sqrt(1-80)]/8
x=[1+-sqrt(-79)]/8
Two solutions:
x=[1+sqrt(-79)]/8 and
x=[1-sqrt(-79)]/8

Answer: Options C. x=[1+sqrt(-79)]/8 and E. x=[1-sqrt(-79)]/8