How do you find the zeros, real and imaginary, of #y=-x^2+4x+12# using the quadratic formula?

2 Answers
Apr 21, 2018

Answer:

See a solution process below:

Explanation:

We can use the quadratic equation to solve this problem:

The quadratic formula states:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(-1)# for #color(red)(a)#

#color(blue)(4)# for #color(blue)(b)#

#color(green)(12)# for #color(green)(c)# gives:

#x = (-color(blue)(4) +- sqrt(color(blue)(4)^2 - (4 * color(red)(-1) * color(green)(12))))/(2 * color(red)(-1))#

#x = (-color(blue)(4) +- sqrt(16 - (-48)))/(-2)#

#x = (-color(blue)(4) +- sqrt(16 + 48))/(-2)#

#x = (-color(blue)(4) +- sqrt(64))/(-2)#

#x = (-color(blue)(4) - 8)/(-2)# and #x = (-color(blue)(4) + 8)/(-2)#

#x = (-12)/(-2)# and #x = (4)/(-2)#

#x = 6# and #x = -2#

Apr 21, 2018

Answer:

#f(x)=a x^2+b x +c#
#x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} #

Explanation:

#f(x)=-x^2+4 x+12=0#
#a=-1, b=4, c=12#

#x_{1}=\frac{-4 + \sqrt{4^2-4*(-1)*12}}{-2}=#

#=\frac{-4 + \sqrt{16+48}}{-2}=(-4 + \sqrt(64))/(-2)=(-4+8)/(-2)=-2#

#x_{2}=\frac{-4 - \sqrt{4^2-4*(-1)*12}}{-2}=#

#=\frac{-4 - \sqrt{16+48}}{-2}=(-4 - \sqrt(64))/(-2)=(-4-8)/(-2)=6#

In this case all zeros are real.

A polynomial of degree n has n zeros. Imaginary zeros are always in conjugate pairs.
#f(x)=x^2+10 x+169#
#x^2+10 x+169=0#
#a=1, b=10, c=169#
#x_{1,2}=\frac{-10\pm\sqrt{10^2-4*1*169}}{2}=#
#=\frac{-10\pm\sqrt{100-676}}{2}=#

#=\frac{-10\pm\sqrt{-576}}{2}#

Remember: #i*i=-1, \sqrt(-1)=i#
#\sqrt{-576}=\sqrt{-2^6*3^2}=2^3*3*i#

#x_{1,2}=\frac{-10\pm\sqrt{-576}}{2}=#
#=\frac{-10\pm 2^3*3*i}{2}=#
#=-5\pm 2^2*3*i=-5\pm 12*i#
#=># since complex zeros are always in pairs, a polynomial of degree #2# has or real or complex zeros.

( a polynomial of degree #3# has:
or #3# real zeros
or #1# real and #2# complex zeros.)

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