Question

Find p if the quadratic equation 4x^2 + (p-2)x + 1 = 0 has equal rooots.

Answer 1

@GirlgoyleH. @ganeshie8 @Gabylovesyou @ilovewolf

Answer 2

@SolomonZelman

Answer 3

@Nnesha help kar lo ess ki...

Answer 4

im sorry im horrid at math

Answer 5

A quadratic equation will have equal roots when the discriminant is zero.

Answer 6

4x^2 + (p-2)x + 1 = 0
Discriminant = b^2 - 4ac = 0 or b^2 = 4ac
That is, (p-2)^2 = 4*4*1 = 16
(p-2)^2 - 16 = 0
(p-2)^2 - 4^2 = 0
(p-2+4)(p-2-4) = 0
(p+2)(p-6) = 0
p = -2, +6

Answer 7

Vieta's Formulas?! @ganeshie8

Answer 8

Please someone help I'm totally confused. :(

Answer 9

im got two answer p=7 or p=6

Answer 10

ok, question, if instead of p-2 you just had an x there, could you find x?

Answer 11

there would be two values you should find for x

Answer 12

@AaronAndyson

Answer 13

Notice that in @aum's problem, he simplify reinstated the quadratic formula by comparison: \[\color{red}ax^2+\color{blue}bx+\color{orange}c=0 \]\[\color{red}4x^2 +\color{blue}{(p-2)}x+\color{orange}1=0\]The discriminant of the quadratic formula tells you not only how many roots there are, but also helps you find out what they are, so now just compare your equation to the quadratic, plug it into the formula for the discriminant and solve to find your roots :)

Answer 14

The discriminant is stated as: \(b^2-4ac\)

Answer 15

I did not get what you said @FibonacciChick666

Answer 16

What ? @Jhannybean

Answer 17

ok, if you have \[4x^2 + rx + 1 = 0\] can you find r? Sorry, bout the x, I didn't realize that was your variable

Answer 18

?
What
?

Answer 19

You're not even trying, it seems.

Answer 20

can you factor \[4x2+rx+1=0\] please

Answer 21

sorry, \[4x^2+rx+1=0\]

Answer 22

@Jhannybean the thing I'm doing this after a very long period of time so it seems like I have forgotten every thing.

Answer 23

So you're unfamiliar with factoring complex quadratics?

Answer 24

It can be said like that.

Answer 25

4x^2 + (p-2)x + 1 = 0
divide by 4:
x^2 + (p-2)/4*x + 1/4 = 0 --- (1)
Assume the equal roots are r and r
Then (x-r)(x-r) = 0
x^2 - 2rx + r^2 = 0 ----- (2)

From (1) and (2): x^2 + (p-2)/4*x + 1/4 = x^2 - 2rx + r^2
Equate the corresponding coefficinets:
(p-2)/4 = 2r ---- (3)
r^2 = 1/4 implies 4r^2 = 1 ---- (4)
Square (3): (p-2)^2 / 16 = 4r^2
From (4): (p-2)^2 / 16 = 1
(p-2)^2 = 16
Take square root:
p-2 = +4 or p-2 = -4
p = 6 or p = -2

Answer 26

There is a negative sign missing above but it should not affect the final answer
(p-2)/4 = -2r ---- (3) (-2r and nor 2r)

Answer 27

pfff nvm