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655-705 (Hard)|   Algebra|      
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Bunuel
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If the difference between the roots of the equation x^2 + ax + 1 = 0 16 Mar 2020, 03:16
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55% (hard)

Question Stats:

65% (02:17) correct 35% (02:18) wrong based on 452 sessions

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If the difference between the roots of the equation \(x^2 + ax + 1 = 0\) is less than √5, then the set of possible values of \(a\) is

A. \(a < -3\)

B. \(-3 < a < 3\)

C. \(3 < a < 5\)

D. \(5 < a < 7\)

E. \(a > 7\)


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B
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If the difference between the roots of the equation x^2 + ax + 1 = 0 16 Mar 2020, 04:18
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Bunuel
If the difference between the roots of the equation \(x^2 + ax + 1 = 0\) is less than √5, then the set of possible values of \(a\) is

A. \(a < -3\)
B. \(-3 < a < 3\)
C. \(3 < a < 5\)
D. \(5 < a < 7\)
E. \(a > 7\)
Show more

Formula: For the quadratic equation \(ax^2 + bx + c=0\), Sum of the roots = \(\frac{−b}{a}\) & Product of the roots = \(\frac{c}{a}\)

Let the roots of \(x^2 + ax + 1 = 0\) be \(m\) & \(n\) respectively
--> \(m + n = \frac{-a}{1} = -a\) & \(mn = 1\)

Given, \(m - n < √5\)
--> \((m - n)^2 < 5\)
--> \((m + n)^2 - 4mn < 5\)
--> \((-a)^2 - 4 < 5\)
--> \(a^2 < 9\)
--> \(a^2 < 3^2\)

Formula: If \(x^2 < a^2\), Solution is \(-a < x < a\)
--> Solution is \(-3 < a < 3\)

Option B
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If the difference between the roots of the equation x^2 + ax + 1 = 0 16 Mar 2020, 14:09
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[-B+{(B^2-4AC)^1/2}]/2A
x^2+ax+1=0 B=a, A=1, C=1
[-a+{(a^2-4)^1/2}]/2 and [-a-{(a^2-4)^1/2}]/2 are two roots, subtracting the two roots
[-a+{(a^2-4)^1/2}]/2 - [-a-{(a^2-4)^1/2}]/2 =2{(a^2-4)^1/2}<5^1/2 .squaring both sides
(a^2-4)<5
a^2<9
-3<a<+3 answer
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If the difference between the roots of the equation x^2 + ax + 1 = 0 03 Jun 2020, 11:24
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If the difference between the roots of the equation x^2 + ax + 1 = 0 05 Jun 2020, 00:48
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Dillesh4096
Bunuel
If the difference between the roots of the equation \(x^2 + ax + 1 = 0\) is less than √5, then the set of possible values of \(a\) is

A. \(a < -3\)
B. \(-3 < a < 3\)
C. \(3 < a < 5\)
D. \(5 < a < 7\)
E. \(a > 7\)

Formula: For the quadratic equation \(ax^2 + bx + c=0\), Sum of the roots = \(\frac{−b}{a}\) & Product of the roots = \(\frac{c}{a}\)

Let the roots of \(x^2 + ax + 1 = 0\) be \(m\) & \(n\) respectively
--> \(m + n = \frac{-a}{1} = -a\) & \(mn = 1\)

Given, \(m - n < √5\)
--> \((m - n)^2 < 5\)
--> \((m + n)^2 - 4mn < 5\)
--> \((-a)^2 - 4 < 5\)
--> \(a^2 < 9\)
--> \(a^2 < 3^2\)

Formula: If \(x^2 < a^2\), Solution is \(-a < x < a\)
--> Solution is \(-3 < a < 3\)

Option B
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Hi

Can you explain to me why -4mn?

Thanks so much
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If the difference between the roots of the equation x^2 + ax + 1 = 0 14 Jul 2020, 02:25
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Dillesh4096
Bunuel
If the difference between the roots of the equation \(x^2 + ax + 1 = 0\) is less than √5, then the set of possible values of \(a\) is

A. \(a < -3\)
B. \(-3 < a < 3\)
C. \(3 < a < 5\)
D. \(5 < a < 7\)
E. \(a > 7\)

Formula: For the quadratic equation \(ax^2 + bx + c=0\), Sum of the roots = \(\frac{−b}{a}\) & Product of the roots = \(\frac{c}{a}\)

Let the roots of \(x^2 + ax + 1 = 0\) be \(m\) & \(n\) respectively
--> \(m + n = \frac{-a}{1} = -a\) & \(mn = 1\)

Given, \(m - n < √5\)
--> \((m - n)^2 < 5\)
--> \((m + n)^2 - 4mn < 5\)
--> \((-a)^2 - 4 < 5\)
--> \(a^2 < 9\)
--> \(a^2 < 3^2\)

Formula: If \(x^2 < a^2\), Solution is \(-a < x < a\)
--> Solution is \(-3 < a < 3\)

Option B
Show more


Hi guys VeritasKarishma chetan2u :)

i know there is this formula for finding roots of quadratic equation \(\frac{-b+\sqrt{D}}{2a }\) and \(\frac{-b-\sqrt{D}}{2a }\)

I wonder what is the difference between formula i mentioned above and this one ---- > For the quadratic equation \(ax^2 + bx + c=0\), Sum of the roots = \(\frac{−b}{a}\) & Product of the roots = \(\frac{c}{a}\)

are they interchangeable ? And in which case it is best to use either of them ?

would appreciate your feedback :)
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If the difference between the roots of the equation x^2 + ax + 1 = 0 16 Jul 2020, 04:26
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dave13
Dillesh4096
Bunuel
If the difference between the roots of the equation \(x^2 + ax + 1 = 0\) is less than √5, then the set of possible values of \(a\) is

A. \(a < -3\)
B. \(-3 < a < 3\)
C. \(3 < a < 5\)
D. \(5 < a < 7\)
E. \(a > 7\)

Formula: For the quadratic equation \(ax^2 + bx + c=0\), Sum of the roots = \(\frac{−b}{a}\) & Product of the roots = \(\frac{c}{a}\)

Let the roots of \(x^2 + ax + 1 = 0\) be \(m\) & \(n\) respectively
--> \(m + n = \frac{-a}{1} = -a\) & \(mn = 1\)

Given, \(m - n < √5\)
--> \((m - n)^2 < 5\)
--> \((m + n)^2 - 4mn < 5\)
--> \((-a)^2 - 4 < 5\)
--> \(a^2 < 9\)
--> \(a^2 < 3^2\)

Formula: If \(x^2 < a^2\), Solution is \(-a < x < a\)
--> Solution is \(-3 < a < 3\)

Option B


Hi guys VeritasKarishma chetan2u :)

i know there is this formula for finding roots of quadratic equation \(\frac{-b+\sqrt{D}}{2a }\) and \(\frac{-b-\sqrt{D}}{2a }\)

I wonder what is the difference between formula i mentioned above and this one ---- > For the quadratic equation \(ax^2 + bx + c=0\), Sum of the roots = \(\frac{−b}{a}\) & Product of the roots = \(\frac{c}{a}\)

are they interchangeable ? And in which case it is best to use either of them ?

would appreciate your feedback :)
Show more

These are different formulas.

Roots are \(\frac{-b+\sqrt{D}}{2a }\) and \(\frac{-b-\sqrt{D}}{2a }\)

Sum of roots = -b/a
\((\frac{-b+\sqrt{D}}{2a })+(\frac{-b-\sqrt{D}}{2a }) = -b/a \)

Product of roots = c/a
\((\frac{-b+\sqrt{D}}{2a }) * (\frac{-b-\sqrt{D}}{2a }) = c/a \)
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If the difference between the roots of the equation x^2 + ax + 1 = 0 13 Oct 2025, 08:49
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