Find the x-intercepts (if any) for the graph of the quadratic function.? #6x^2 +12x+5=0#

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2 Answers
AltairSafir
Mar 22, 2018

Just apply the formula #x=(-b(+)or(-)(b^2-4*a*c)^(1/2))/(2*a)#
where the quadratic function is # a*x^2+b*x+c=0#

Explanation:

In your case:
#a=6#
#b=12#
#c=5#
#x_(1)=(-12+(12^2-4*6*5)^(1/2))/(2*6)=-0.59#
#x_2=(-12-(12^2-4*6*5)^(1/2))/(2*6)=-1.40#

Anish M.
Mar 22, 2018

#-0.5917# and #-1.408#

Explanation:

The x intercepts are basically the points where the line touches the x-axis. On the x-axis, the y co-ordinate is always zero so now we find values of x for which #6x^2 +12x + 5# = 0.

This is a quadratic equation and we can solve this using the quadratic formula:
#x# = #(-b+-sqrt(b^2-4*a*c))/(2*a)#

Now, for #6x^2+12x+5#,
a=6. b=12, c=5.
On substituting the values in the formula, we get

#x#= #(-12+-sqrt(12^2-4*6*5))/(2*6)#

#=# #(-12+-sqrt(144-120))/(12)#

#=# #(-12+-sqrt(24))/(12)#

This gives us the two values as #-0.5917# and #-1.408#

Hence the two #x# intercepts for the given equation are #-0.5917# and #-1.408#.

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