Given that the curve y=ax^2 + 3ax + a + b, where a≠0, intersects the x-axis at exactly one point, find b in terms of a.

Question

anonymous · Answer

The answer given is 5/4 a

anonymous · Answer

But when I tried working it out I got 1/4 a

anonymous · Answer

$$a= a, b=3a,c=a+b$$$$3a ^{2}-4(a)(a+b)=0$$$$3a ^{2}-4a ^{2}+4ab=0$$$$-a ^{2}+4ab=0$$$$4ab=a ^{2}$$$$4b=a$$$$b=\frac{ 1 }{ 4 }a$$

anonymous · Answer

This is what I did

anonymous · Answer

erm ill try and help let me have a look.

anonymous · Answer

alright thanks!

anonymous · Answer

@JamesWolf

anonymous · Answer

trying to solve it with the quadratic equation, which is not your method gives me -3/2, so I'm abit stuck really, sorry

anonymous · Answer

oh alright then.... Thanks anyways.

anonymous · Answer

Im annoyed. I dont understand why it didnt work

anonymous · Answer

if it only crosses the x axis once, then that means it looks like this |dw:1371590678439:dw|

dumbcow · Answer

i got it

(3a)^2 = 9a^2  not 3a^2

dumbcow · Answer

other than that @BelleFlower your steps are correct

anonymous · Answer

or some flip of that. so if you use the quadratic equation 

(-b +- sqrt(b^2 -4ac))/2a

the square root, sqrt(), should equal 0. since there is only one x intercept. So were left with 

-3a/2a = -3/2

@dumbcow, where have I gone wrong here?