Question

find the zeros of 2x^2 - 16x + 27 using the quadratic formula. be sure to simplify the expression. Can someone please show me how this is done.

Answer 1

For a quadratic of the form
`f(x)=ax^2+bx+c`, we have the quadratic formula

`x=(-b \pm √(b^2 -4ac) )/(2a)`,
where a is the coefficient (number before the variable) of the squared term, b is the coefficient of the linear term, and c is the constant term.

So, given
`2x^2-16x+27=0`, we can get that
`a=2, \ b=-16`, and
`c=27`. We substitute these numbers into the quadratic formula above.


`x=(-(-16) \pm √((-16)^2 -4(2)(27)) )/(2(2))`


`x=(16 \pm √((256 -216)) )/(4)`


`x=(16 \pm √(40) )/(4)`


`x=(16 \pm 2√(10) )/(4)`


`x=4+ (√(10))/(2), \ x=4- (√(10))/(2)`

This is our final answer.

If you've never seen the quadratic formula, you can derive it by completing the square for the general form of a quadratic. Note that the
`\pm` symbol (read: plus or minus) represents the two possible distinct solutions, except for zero under the radical, which gives only one solution.