How do you find the zeros, real and imaginary, of #y= -x^2-55x+37 # using the quadratic formula?

2 Answers
May 19, 2017

Answer:

#" "x~~0.665 and x~~-55.665# to 3 decimal places

#x=-55/2+-sqrt(3173)/2# as exact values

Explanation:

#color(blue)("Preamble")#
When I first came across the quadratic formula I decided that I really had to commit it to memory. So I wrote the thing out every time I answered a question of this type. I did it so often that it is really 'burnt' into my memory. This is one of those equations that is really worth remembering if you can.

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#color(blue)("Answering your question")#

The coefficient of #x^2# is negative (-1) so the graph is of form #nn#

The constant of #37# is the y-intercept indicating that the graph has to cross the x-axis. Thus the roots are real numbers

Compare to standard form#" "y=ax^2+bx+c#

Setting #y=0# gives #x=(-b+-sqrt(b^2-4ac))/(2a)#

Where #a-1"; "b=-55"; "c=37#

#x=(+55+-sqrt((-55)^2-4(-1)(37)))/(2(-1))#

#x=-55/2+-sqrt(3173)/2#

Note that as we are applying plus or minus to the root it does not matter that the denominator is negative. So write it as positive just to make things tidy.

The only whole number factors of 3173 are: 1 and 3173; 19 and 167. so we can not simplify the root any further than it is.

Thus the exact answer is: #" "x=-55/2+-sqrt(3173)/2#

Approximate answers are:

#" "x~~0.665 and x~~-55.665# to 3 decimal places
Tony B

May 19, 2017

Answer:

Zeros are #-55/2+sqrt(3173)/2# and #-55/2-sqrt(3173)/2#

Explanation:

For a quadratic polynomial #y=ax^2+bx+c#,

quadratic formula gives the zeros as #(-b+-sqrt(b^2-4ac))/(2a)#

Hence for #y=-x^2-55x+37#, we have #a=-1, b=-55# and #c=37#

and hence zeros are #(-(-55)+-sqrt((-55)^2-4xx(-1)xx37))/(-2)#

= #(55+-sqrt(3025+148))/(-2)#

= #-55/2+-sqrt(3173)/2#

Note #3173=167xx19# and hence no square root can be taken out.