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Let d and e denote the solutions of 2x^2 + 3x-5=0. What is the value o

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Let d and e denote the solutions of 2x^2 + 3x-5=0. What is the value o  [#permalink]

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New post 19 Mar 2019, 02:03
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

86% (01:24) correct 14% (01:48) wrong based on 35 sessions

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Re: Let d and e denote the solutions of 2x^2 + 3x-5=0. What is the value o  [#permalink]

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New post 19 Mar 2019, 02:15
Bunuel wrote:
Let d and e denote the solutions of \(2x^{2}+3x-5=0\). What is the value of (d-1)(e-1)?

(A) -5/2
(B) 0
(C) 3
(D) 5
(E) 6



We can solve \(2x^{2}+3x-5=0\) to get solutions as -5/2 & 1.
Value = (d-1)(e-1)

Since (d-1) or (e-1) will be equal to 0 as one of the root is 1.
So value of (d-1)(e-1) = (-5/2 - 1)(1-1) = 0 (Option B)
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Let d and e denote the solutions of 2x^2 + 3x-5=0. What is the value o  [#permalink]

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New post 19 Mar 2019, 06:03
Bunuel wrote:
Let d and e denote the solutions of \(2x^{2}+3x-5=0\). What is the value of (d-1)(e-1)?

(A) -5/2
(B) 0
(C) 3
(D) 5
(E) 6


solve
\(2x^{2}+3x-5=0\)
we get
-5/2 * 1
so
(+1-1)*(-5/2-1)
0
IMO b
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Let d and e denote the solutions of 2x^2 + 3x-5=0. What is the value o  [#permalink]

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New post 19 Mar 2019, 18:37
Using Viete's theorem:

\(2x^2 + 3x-5=0\)

\(Sum: \frac{-b}{a}\)

\(\frac{-3}{2}\)


\(Prod: \frac{c}{a}\)

\(\frac{-5}{2}\)


Given: \((d-1)(e-1)\) = \(de\) \(-d -e\) \(+1\) =

\(de\) \(- (d+e\)) \(+1\)

\(\frac{-5}{2} - (\frac{-3}{2}) + 1\)

\(\frac{-5}{2} + \frac{3}{2}\) \(+ 1\)

\(\frac{-2}{2} = -1\) \(+ 1\) \(= 0\)

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Re: Let d and e denote the solutions of 2x^2 + 3x-5=0. What is the value o  [#permalink]

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New post 21 Mar 2019, 18:02
Bunuel wrote:
Let d and e denote the solutions of \(2x^{2}+3x-5=0\). What is the value of (d-1)(e-1)?

(A) -5/2
(B) 0
(C) 3
(D) 5
(E) 6


Solving for x, we have:

(2x + 5)(x - 1) = 0

2x + 5 = 0

2x = -5

x = -5/2

or

x = 1

Thus,

(d - 1)(e - 1) = (-2.5 - 1)(1 - 1) = 0

Alternate solution:

We can use the following fact:

If r and s are solutions (or roots) of the quadratic equation ax^2 + bx + c = 0 (where a ≠ 0), then r + s = -b/a and rs = c/a.

Here, since d and e are solutions of 2x^2 + 3x - 5 = 0, then d + e = -3/2 and de = -5/2. Therefore, we have:

(d - 1)(e - 1) = de - d - e + 1 = de - (d + e) + 1 = -5/2 - (-3/2) + 1 = 0

Answer: B
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Re: Let d and e denote the solutions of 2x^2 + 3x-5=0. What is the value o   [#permalink] 21 Mar 2019, 18:02
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