How do you find the value of the discriminant and determine the nature of the roots #-4m^2-4m+5#?

4 Answers
Sep 8, 2017

Answer:

#Delta = 96#, and there are two real roots.

Explanation:

In a quadratic function written in the form
#ax^2 + bx +c#
the discriminant, also called delta is known as follow:
#Delta# = #b^2 -4ac#

In the case of #-4m^2 -4m +5# the x's have been replaced with m's, but the formula still applies. We'll take #a=-4#, #b=-4#, and #c=5#.
Then using the formula for the discriminant...

#Delta = b^2 -4ac#
#=(-4)^2 - 4(-4)(5)#
#=16+80#
#=96#

So #Delta = 96#
When they ask about the 'nature of the roots' they what this basically means is, how many are there? You can know this from the descriminant.
When #Delta > 0#, there's two distinct, separate real roots.
When #Delta = 0#, there's only one real root, which we call a "double root".
When #Delta < 0#, there are no real roots.
(As for why they're called 'real' roots, you don't need to worry about that now)

So in this case, when #Delta = 96# it is also #>0#, meaning there are two real roots.

Sep 8, 2017

Answer:

Discriminant: #Delta=96#
implying #2# Real roots

Explanation:

Given a quadratic of the form:
#color(white)("XXX")color(red)am^2+color(blue)bm+color(brown)c color(white)("xxxx")#..see note below
the discriminant is
#color(white)("XXX")Delta=color(blue)b^2-4color(red)acolor(brown)c#
with
#color(white)("XXX"){(0" Real roots (2 Complex roots)","if", Delta < 0), (1" Real root","if",Delta = 0), (2" Real roots (0 Complex roots)","if",Delta > 0) :}#

For the given example: #color(red)(-4)m^2color(blue)(-4)m+color(brown)5#
we have
#color(white)("XXX")Delta=(color(blue)(-4))^2-4 * (color(red)(-4)) * color(brown)(5)#

#color(white)("XXXX")=16+80#

#color(white)("XXXX") > 0#

So this expression has #2# roots

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Note:
The most common problem with this type of question is that we are typically accustomed to seeing this with #x# used as the variable,
but the variable can be any letter (in this case it happens to be #m#).

Sep 8, 2017

Answer:

#"2 real irrational roots"#

Explanation:

#Delta=b^2-4aclarrcolor(blue)" discriminant"#

#• " if "Delta>0" then 2 real irrational roots"#

#• " if "Delta>0" and a perfect square"#

#"then 2 real rational roots"#

#• " if "Delta=0" then real rational equal roots"#

#• " if "Delta<0" then 2 complex conjugate roots"#

#-4m^2-4m+5#

#"with "a=-4,b=-4,c=5#

#Delta=b^2-4ac=16+80=96#

#rArr"2 real and irrational roots"#

#color(blue)"As a check"#

#"solve for m using the "color(blue)"quadratic formula"#

#m=(4+-sqrt96)/(-8)=(4+-4sqrt6)/(-8)#

#rArrm=-1/2+-1/2sqrt6larr" 2 real irrational roots"#

Sep 8, 2017

Answer:

Two real solutions , #m ~~ -1.72 , m ~~ 0.72#

Explanation:

#-4m^2-4m+5 =0 # . Comparing with the standard quadratic

equation #ax^2+bx+c =0 # , we get, # a= -4 , b = -4 ,c=5#

Discriminant #D = (b^2-4ac) = 16+80=96 #

If discriminant is positive, we get two real solutions, if it is zero we

get just one solution, and if it is negative we get complex solutions.

Here discriminant is positive, so we get two real solutions.

# m= (-b+- sqrtD)/(2a) = (4 +- sqrt 96)/ -8 # or

#m = - (4+-4sqrt6)/8 = (-4 (1+-sqrt6))/8 # or

# m= - (1+-sqrt6)/2 = -1/2 +- sqrt6/2# or

#m ~~ -1.72(2dp) , m ~~ 0.72(2dp) # [Ans]