Finding the coefficients in a quadratic equation when you know the two roots

**jonnygill** · March 11th 2011, 04:33 PM

The problem is...

Find a, b, c if...

$P(x)=ax^2+bx+c$ AND $P(3+4i)=0$ and $P(3-4i)=0$

So, we know that 3+4i and 3-4i are two roots (imaginary).

The way I approached this problem is I looked at the quadratic formula and I set a to a value such that the denominator equals 1.

So, I let a=.5

The formula becomes $-b\pm\sqrt{b^2-2c}$

I figured b must equal -3 so that -b=3.

I also figured that c must equal 12.5 so that the radical becomes $\sqrt{-16}$

So i got...

a=.5
b=-3
c=12.5

The answer in the back of the book gives an answer of...

a=1
b=-6
c=25

Basically, they doubled all the coefficients. Are both answers acceptable? If only the books answer is acceptable, how could I solve this problem and arrive at the answer the book provided?

Thanks.

**Prove It** · March 11th 2011, 05:15 PM

If $\displaystyle 3 + 4i$ and $\displaystyle 3 - 4i$ are roots of $\displaystyle P(x) = ax^2 + bx + c$ , then

$\displaystyle a(3 + 4i)^2 + b(3 + 4i) + c = 0$ and $\displaystyle a(3 - 4i)^2 + b(3 + 4i) + c = 0$ .

Simplifying equation 1: $\displaystyle a(-7 + 24i) + 3b + 4bi + c = 0$

$\displaystyle -7a + 24ai + 3b + 4bi + c = 0$

$\displaystyle -7a+3b+c + (24a + 4b)i = 0$

Therefore $\displaystyle -7a + 3b + c = 0$ and $\displaystyle 24a + 4b = 0$ when you equate real and imaginary parts.

Now do the same with Equation 2, and solve the resulting system for $\displaystyle a,b,c$ .

**TheChaz** · March 11th 2011, 05:39 PM

Wow. That's intense!
x = 3 + 4i
x - 3 = 4i
Square both sides...
x^2 - 6x + 9 = -16
x^2 - 6x + 25.

**LoblawsLawBlog** · March 11th 2011, 07:27 PM

Is that exactly how the problem is worded? Because
$a\bigl(x-(3+4i)\bigr)\bigl(x-(3-4i)\bigr)$
will have the same zeros for any value of a.

Your solution works, as does the book's.

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