Question

For a quadratic equation in the form ax^2+bx+c=0, find the other root if one root is 8 and a=1 and b=-4

Answer 1

x^2 -4x + c = 0

and (x - 8)(x + r) = 0 where r = other root

sum of the roots = -b/a = --4/ 1 = 4
so 8 + r = 4
r = -4

other root = -4

Answer 2

See, here a = 1, b = -2 and let c = c

So you know:
\[D = b^2 - 4ac\]

D = 16 - 4c

And roots can be found as:

\[x = (-b \pm \sqrt{D})/2a\]

It is given one of the root is 8:

so,
\[8 = (-(-4) + \sqrt{(16-4c)})/2\]

\[8 = 2 + \sqrt{(4-c)}\]

\[4 - c = 6^2\]

c = -32

So, put this value in D..

\[D = 16 + 128 = 144\]

So other root will be:

\[x = (4 - \sqrt{D})/2\]

Put the value of D and get the other value of x...

Answer 3

So what if one root =1/2 a=-1 and c=6 ?

Answer 4

Then use the product as Diajiones will explain it to you...

Answer 5

@waterineyes the answer you gave me, is not right. so

Answer 6

i dont know if you've done this in class yet by i used the following identities

if A and B are the roots of the equation ax^2 + bx + c = 0
then

A + B = -b/a

and AB = c /a

Answer 7

Huh?

Answer 8

a = -1 c = -6 and 1 root is 1/2

then 1/2 * r = c/a = -6/-1 = 6

r/2 = 6

solve for r to get other root

Answer 9

Did you solve it for x??? @Pamelaa23

Answer 10

Im lost!

Answer 11

can you solve this equation for ;