Question

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

Answer 1

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`

Answer 2

`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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