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# If m and n are the roots of the quadratic equation x2 - (2 root 5)x -

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Joined: 02 Sep 2009
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If m and n are the roots of the quadratic equation x2 - (2 root 5)x -  [#permalink]

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13 Mar 2016, 09:58
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Difficulty:

45% (medium)

Question Stats:

66% (01:57) correct 34% (02:21) wrong based on 127 sessions

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If m and n are the roots of the quadratic equation $$x^2 - (2 \sqrt 5)x - 2 = 0$$, the value of $$m^2 + n^2$$ is:

A. 18
B. 20
C. 22
D. 24
E. 32

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Posts: 388
Location: India
GMAT 1: 760 Q50 V44
Re: If m and n are the roots of the quadratic equation x2 - (2 root 5)x -  [#permalink]

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15 Mar 2016, 05:17
1
5
Bunuel wrote:
If m and n are the roots of the quadratic equation $$x^2 - (2 \sqrt 5)x - 2 = 0$$, the value of $$m^2 + n^2$$ is:

A. 18
B. 20
C. 22
D. 24
E. 32

In the equation ax^2 + bx + c =0
Sum of roots = -b/a
Product of roots = c/a

Sum of roots (m + n) = $$(2 \sqrt 5)$$
Product of roots (mn) = -2

$$m^2 + n^2$$ = $$(m + n)^2 - 2mn$$ = 20 + 4 = 24
Option D
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Re: If m and n are the roots of the quadratic equation x2 - (2 root 5)x -  [#permalink]

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13 Mar 2016, 10:10
2
Bunuel wrote:
If m and n are the roots of the quadratic equation $$x^2 - (2 \sqrt 5)x - 2 = 0$$, the value of $$m^2 + n^2$$ is:

A. 18
B. 20
C. 22
D. 24
E. 32

Product of root, MN = C/A = -2/1 = -2
Sum of root, M+N = -B/A = -(-2\sqrt{5}/1 )= 2\sqrt{5}

(M+N)^2 = M^2 + N^2 + 2MN = 4*5 = 20

2MN = -2(2) = -4

M^2 + N^2 = 20 + 4 = 24

IMO .D
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Re: If m and n are the roots of the quadratic equation x2 - (2 root 5)x -  [#permalink]

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15 Mar 2016, 01:29
Here D = B^2-AC = 28
so the roots are 2√5 +√28/2 and 2√5 -28/2
hence M^2 +N^2 => 28
i.e. D
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Re: If m and n are the roots of the quadratic equation x2 - (2 root 5)x -  [#permalink]

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15 Jun 2016, 03:39
stonecold wrote:
Here D = B^2-AC = 28
so the roots are 2√5 +√28/2 and 2√5 -28/2
hence M^2 +N^2 => 28
i.e. D

Apology for seeing a typo error It should be 24 and not 28.
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Re: If m and n are the roots of the quadratic equation x2 - (2 root 5)x -  [#permalink]

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15 Jun 2016, 04:12
Bunuel wrote:
If m and n are the roots of the quadratic equation $$x^2 - (2 \sqrt 5)x - 2 = 0$$, the value of $$m^2 + n^2$$ is:

A. 18
B. 20
C. 22
D. 24
E. 32

m+n = -{- (2 \sqrt 5)}
mn = -2

Squaring 1st equation we get

$$m^2+n^2+2mn = 20$$

$$m^2+n^2= 20 - 2 *(-2)$$

$$m^2+n^2= 20 + 4$$ = 24

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Joined: 12 Sep 2017
Posts: 246
Re: If m and n are the roots of the quadratic equation x2 - (2 root 5)x -  [#permalink]

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25 Feb 2019, 19:53
TeamGMATIFY wrote:
Bunuel wrote:
If m and n are the roots of the quadratic equation $$x^2 - (2 \sqrt 5)x - 2 = 0$$, the value of $$m^2 + n^2$$ is:

A. 18
B. 20
C. 22
D. 24
E. 32

In the equation ax^2 + bx + c =0
Sum of roots = -b/a
Product of roots = c/a

Sum of roots (m + n) = $$(2 \sqrt 5)$$
Product of roots (mn) = -2

$$m^2 + n^2$$ = $$(m + n)^2 - 2mn$$ = 20 + 4 = 24
Option D

Hello!

Could someone please clarify to me why $$m^2 + n^2$$ is the same as:

$$m^2 - 2mn + n^2$$

Kind regards!
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Re: If m and n are the roots of the quadratic equation x2 - (2 root 5)x -  [#permalink]

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02 Mar 2019, 10:11
Bunuel wrote:
If m and n are the roots of the quadratic equation $$x^2 - (2 \sqrt 5)x - 2 = 0$$, the value of $$m^2 + n^2$$ is:

A. 18
B. 20
C. 22
D. 24
E. 32

For any quadratic equation of the form x^2 + bx + c = 0, if r and s are the roots of the equation, then r + s = -b and rs = c. So here we have:

m + n = 2√5 and mn = -2

Squaring the first equation, we have:

(m + n)^2 = (2√5)^2

m^2 + 2mn + n^2 = 20

Since mn = -2, we have:

m^2 + 2(-2) + n^2 = 20

m^2 - 4 + n^2 = 20

m^2 + n^2 = 24

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Re: If m and n are the roots of the quadratic equation x2 - (2 root 5)x -   [#permalink] 02 Mar 2019, 10:11
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