Determine a value of a such that one root of the equation ax^2+x-1=0 is five times the other root. Explain.?

Let β and 5β be the two roots of the equation ax² + x - 1 = 0

The sum of the two roots:
β + 5β = -1/a
Then, 6aβ = -1
36a²β² = 1 …… [1]

The product of the two roots:
β * 5β = -1/a
5aβ² = -1 …… [2]

[1]/[2]:
36a²β² / (5aβ²) = 1 / (-1)
36a/5 = -1
a = -5/36

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Check:
(-5/36)x² + x - 1 = 0
[(-5/36)x² + x - 1] * (-36) = 0
5x² - 36x + 36 = 0
(5x - 6)(x - 6) = 0
x = 6/5 or x = 6

The root 6 is five times the other root 6/5.

:--
[ (--1 ) -- √ ( 1 + 4a ) ] = 5 [ (-1) + √ ( 1 + 4a ) ]
4 = 2 √ ( 1 + 4a )
4 = 1 + 4a
a = 3/4

Let x1, x2 be the roots of ax^2+x-1=0, then
x1x2=-1/a-----(1)
x1+x2=-1/a----(2)
x2=5x1--------(3)
Since
ax1^2+x1-1=0----(4)
ax2^2+x2-1=0
=> from (3), get
25ax1^2+5x1-1=0---(5)
(5)-(4)=>
4x1(6ax1+1)=0=>
x1=-1/(6a), since x1=/=0 for otherwise
a is infinity.
From (1), get
5(-1/6a)^2=-1/a=>
5/(36a^2)=-1/a=>
a=-5/36.

Check: from (2) & (3), we have
LHS=
x1+x2=
-1/(6a)+5[-1/(6a)]=
-6/(6a)=
-1/a=
RHS
valid.

Roots α , 5α
α+5α = -1/a
α(5α)=-1/a
---> α = 6/5
and
a=-5/36

Explain what, exactly? The equation has two roots. One of them is five times the other root. Is that so hard to understand?

<sigh> Try a = 13 / 36