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Mathematics 15 Online
OpenStudy (anonymous):

Click here

OpenStudy (anonymous):

(a^2 -a -2 )x^2 + (a^2 - 4)x + (a^2 -3a +2 ) = 0 is a quadratic equn Does there exist a real value for "x" for which the above equation will be an identity in "a"

OpenStudy (anonymous):

@nincompoop

OpenStudy (anonymous):

@sidsiddhartha

OpenStudy (anonymous):

How to check this ?

OpenStudy (anonymous):

@mathslover

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

everyone is clicking here but no answers yet...

OpenStudy (anonymous):

yes

OpenStudy (loser66):

Let me try once

OpenStudy (loser66):

take discriminant and you have \[\triangle = -3z^4+16a^3-20a^2-16a+32\] if \(\triangle < 0\) --> no solution if \(\triangle =0\) double root if \(\triangle >0\) 2 distinct roots so that we should check whether \(\triangle \geq 0\) I compute and got a = 2 and some number around -0.85.. Hey, mathematicians, tell me whether I can make conclusion from here? Please correct me if I am wrong

OpenStudy (anonymous):

where did u get "z" from

OpenStudy (loser66):

what is z?

OpenStudy (loser66):

oh, that is typo, it's a^4

OpenStudy (anonymous):

Gys really appreciate it if u helped me on a question Medal on this page + Medal on my question page + fan. @adi.arju.2000 Thanks :) P.S the question is; A drum vibrates 180 times in 2 seconds. If the speed of sound is 340m/s, what is the period, frequency and wavelength of the waves produced by the drum? The speed of light is 3 X 10^8m/s. Thanks :')

OpenStudy (anonymous):

Sry for puttin this on a random question :/

OpenStudy (shubhamsrg):

After the expressing the quadratic eqn in the form of f(x) a^2 + g(x) a + h(x) = 0, I think it can be an identity for some value of x , say n, only if f(n) = g(n) = h(n) = 0.

OpenStudy (anonymous):

ok i will see this late ty all for the help

OpenStudy (nipunmalhotra93):

what exactly do you mean by identity in a?

OpenStudy (shubhamsrg):

Perhaps it means that the expression should be true for any value of a.

OpenStudy (nipunmalhotra93):

oh I see... thanks :)

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