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.!