if the ratio of the roots of ax2+bx+c=0 is m, show that (m+1)^2 ac=b^2m
Answers
Step-by-step explanation:
The roots of the equation ax^2 + bx + c = 0 are in the ratio of 2 :3, then what is the relation between a, b and c?
The roots of the equation ax^2 + bx + c = 0 are
x1 =[-b +(b^2 -4ac)^0.5]/2a, and
x2 = [-b -(b^2 -4ac)^0.5]/2a.
Now it says x2:x1::2:3, or
{[-b -(b^2 -4ac)^0.5]/2a}/{[-b +(b^2 -4ac)^0.5]/2a} = 2/3, or
3{[-b -(b^2 -4ac)^0.5]/2a} = 2{[-b +(b^2 -4ac)^0.5]/2a}, or
-3b -3(b^2 -4ac)^0.5 = -2b +2(b^2 -4ac)^0.5, or
-3b +2b = 2(b^2 -4ac)^0.5 + 3(b^2 -4ac)^0.5, or
-b = 5(b^2 -4ac)^0.5, or
-b/5 = (b^2 -4ac)^0.5. Squaring both sides we get
b^2/25 = b^2 - 4ac, or
b^2 = 25(b^2 - 4ac), or
24b^2 = 100ac, or
b^2 =25ac/6, or
b = 5(ac/6)^0.5, which is the relation between a, b and c.
Check: ax^2 + bx + c = 0 can be written as
ax^2 + 5(ac/6)^0.5x + c = 0
x1 = -5(ac/6)^0.5 +[{5(ac/6)^0.5}^2 -4ac]^0.5, or
= -5(ac/6)^0.5 +{25ac/6 - 4ac}^0.5
= -5(ac/6)^0.5 +{25ac - 24ac}^0.5/6^0.5, or
=-5(ac)^0.5/6^0.5 +(ac)^0.5/6^0.5, or
= -4(ac)^0.5/6^0.5
x2= -5(ac/6)^0.5 -[{5(ac/6)^0.5}^2 -4ac]^0.5, or
= -5(ac/6)^0.5 -{25ac/6 - 4ac}^0.5
= -5(ac/6)^0.5 -{25ac - 24ac}^0.5/6^0.5, or
=-5(ac)^0.5/6^0.5 -(ac)^0.5/6^0.5, or
= -6(ac)^0.5/6^0.5
Therefore x1:x2::-4:-6 or 2:3. Proved.
Answer is: b = 5(ac/6)^0.5, which is the relation between a, b and c.
Answer:
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