Question #a0eac

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Answer 1 · 2017-04-01T17:16:17

#a = 1, b = -5n, c = 4n^2#, where #n# = the number

Let #n =# the number, let #4n=# 4 times the number

#(x - n)( x - 4n) = 0#

Distribute: #x^2 - 5nx + 4n^2 = 0#

Compare to #ax^2 +bx + c = 0#:

#a = 1, b = -5n, c = 4n^2#

Check:
Let #n = 5, 4x = 20#
#(x - 5)(x - 20) = 0#
#x^2 - 25x + 100 = 0#
#a = 1, b = -5*5 = -25, c = 4*5^2 = 100#

Let #n = -5, 4x = -20#
#(x + 5)(x + 20) = 0#
#x^2 + 25x + 100 = 0#
#a = 1, b = 5*5 = 25, c = 4*(-5)^2 = 100#

Answer 2 · 2017-04-01T18:09:22

#4b^2=25ac.#

Let #alpha and beta# be the Roots of the given Quadr.

By what is given, we may, assume, #beta=4alpha.....(1).#

We also know that, #alpha+beta=-b/a....(2), &, alpha*beta=c/a...(3).#

#(1) & (2) rArr 5alpha=-b/a, i.e., alpha=-b/(5a)....(4).#

#(1) & (3) rArr 4alpha^2=c/a, i.e., alpha^2=c/(4a)......(5).#

#"Finally, "(4) & (5) rArr c/(4a)={-b/(5a)}^2=b^2/(25a^2).#

#:. 25ac=4b^2,# is the desired relation btwn. #a,b,c.#

Enjoy Maths.!