If the roots of ax^2 + bx + c differ by 3, show that b^2 = 9a^2 + 4ac?

Question

This is what i did but i don't know what i'm doing wrong here, can you check this for me please:

x = (-b +/- √ b^2 - 4ac) / 2a

Xi = (-b +√ b^2 - 4ac) / 2a. Xii = (-b - √ b^2 - 4ac) / 2a

Xi - Xii = 3 since they differ by 3

Xi - Xii = [(-b +√ b^2 - 4ac) / 2a] - [(-b - √ b^2 - 4ac) / 2a]

= (-b +√ b^2 - 4ac) / 2a + (b +√ b^2 - 4ac) / 2a

= (b^2 - 4ac) / 2a, since Xi - Xii = 3 therefore

3 = (b^2 - 4ac) / 2a

6a = b^2 - 4ac

b^2 = 6a + 4ac <<<<

Answer 1

I will start from the midway in your own answer.



Xi - Xii = [(-b +√ b^2 - 4ac) / 2a] - [(-b - √ b^2 - 4ac) / 2a]

= (-b +√ b^2 - 4ac) / 2a + (b +√ b^2 - 4ac) / 2a

Take the common denominator out

= [-b +√ (b^2 - 4ac) + b +√ (b^2 - 4ac)] / 2a

= [√ (b^2 - 4ac) + √ (b^2 - 4ac)] / 2a

= 2 √(b^2 - 4ac) / 2a

= √(b^2 - 4ac) / a



This equals to 3 as per the given fact



So, √(b^2 - 4ac) / a = 3



ie. √(b^2 - 4ac) = 3 a



Square both sides

(b^2 - 4ac) = 9 a^2





So, b^2 = 9a^2 + 4ac . Hence Proved

Answer 2

On your fifth line you have...



(-b +√ b^2 - 4ac) / 2a) + (b +√ b^2 - 4ac) / 2a



But on your next line you have that this simplifies to (b^2 - 4ac)/2a. But actually, the only things that cancel out are the b's (-b + b = 0).



So you're left with...



(√ b^2 - 4ac)/2a + (√ b^2 - 4ac) / 2a

= 2(√b^2 - 4ac)/2a = (√b^2 - 4ac)/a



Now set that equal to 3 and solve for b...

Answer 3

ax^2 + bx + c =0



3 = [-b + (b^2-4ac)^(1/2)] / 2a -

[-b - (b^2-4ac)^(1/2)] / 2a



3 = [2(b^2-4ac)^(1/2)] / 2a



3= [(b^2-4ac)^(1/2)] / a



3a = (b^2-4ac)^(1/2)



9a^2 = b^2 - 4ac



b^2 = 9a^2 + 4ac