inanisumeet
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- Mar 26, 2020
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Suppose that a = kc and b = kd. Expand (a+b)/(c+d) and sqrt((ab)/(cd)).I found a question in math magazine that said
if ratio of roots of ax2+2bx+c=0 is same as ratio of roots of px2+2qx+r=0 , then
(1)2b/ac=q2/pr
(2)b/ac=q/pr
(3)b2/ac=q2/pr
(4)None of these
Answer is given as (3) and the hint is View attachment 25330
How did they make the last statement in above picture ?
Thanks.
How did they make the last statement in above picture ?
yes , in that case statement is valid.Suppose that a = kc and b = kd. Expand (a+b)/(c+d) and sqrt((ab)/(cd)).
yesAre the Greek letter the roots?
I'm not sure exactly how this relates to the original question, but it's true that there are conditions under which the step you asked about is not valid; they don't seem to have stated those conditions, so what they say is not universally valid. (Cubist has mentioned another condition, involving signs.)yes , in that case statement is valid.
If we say 2/3=(-2)/(-3) which is not equal to (2+(-2))/(3+(-3))
I'm not sure exactly how this relates to the original question, but it's true that there are conditions under which the step you asked about is not valid; they don't seem to have stated those conditions, so what they say is not universally valid. (Cubist has mentioned another condition, involving signs.)
In your example here, a/b = c/d is equal to (a+b)/(c+d) when all three ratios are defined; using my approach to prove this, letting a = kb and c = kd we find that (a+b)/(c+d) = ([b(k+1)]/[d(k+1)], which equals b/d as long as k is not -1 -- that is, only when the two given ratios are not -1 as they are in your counterexample. So that condition should be stated.