+0

# Let equation and (a≠c) have the same root . Please determine this root value using a, b, c, d to represent it.

+1
248
3

Let equation $$x^2+ax+b=0$$ and $$x^2+cx+d=0$$    (a≠c) have the same root $$x$$. Please determine this root value using a, b, c, d to represent it.

Sep 24, 2018

#1
+4

we can add the two equations to obtain

$$2x^2 +(a+c)x+(b+d)=0\\ x^2 + \dfrac{a+c}{2}x+\dfrac{b+d}{2}=0$$

$$\text{Now just apply the quadratic formula}\\ r_{1,2} = \dfrac{-\frac{a+c}{2}\pm \sqrt{\left(\frac{a+c}{2}\right)^2 - 2(b+d)}}{2}$$

$$\text{we can clean this up a bit}\\ r_{1,2} = \frac{1}{4} \left(\pm\sqrt{(a+c)^2-8 (b+d)}-a-c\right)$$

$$\text{If there is only a single shared root }r_1 = r_2\\ \text{if there are no shared roots}\\ (a+c)^2 - 8(b+d)<0$$

.
Sep 25, 2018
#2
+2

Why add the two equations ?

Why not simply subtract one from the other ?

Suppose that the first equation has roots p and q and that the second equation has roots p and s,

so that the common root is p.

Then we have

$$\displaystyle (x-p)(x-q)=0 \text{ and }(x-p)(x-s)=0\; .$$

Subtracting the second equation from the first,

$$\displaystyle (x-p)\{(x-q)-(x-s)\}=0\; ,$$

so

$$\displaystyle (x-p)(s-q)=0\; ,$$

showing that  x = p  is the  root of the resulting equation,  $$\displaystyle (\;s \neq q\;)\;.$$

So, going back to the original equations and subtracting one from the other,

$$\displaystyle x(a-c)+(b-d)=0\;,$$

$$\displaystyle x=\frac{d-b}{a-c}\; .$$

Tiggsy

Sep 25, 2018
#3
0

you're right though I'd do it as

$$x^2+a x+b - (x^2+c x+d) = 0\\ (a-c)x + (b-d) = 0 \\ \forall a \neq c \\ x = \dfrac{d-b}{a-c}$$

Rom  Sep 27, 2018