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

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Bunuel
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If m and n are the roots of the quadratic equation x2 - (2 root 5)x - [#permalink] New post   13 Mar 2016, 09:58
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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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D
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TeamGMATIFY
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Re: If m and n are the roots of the quadratic equation x2 - (2 root 5)x - [#permalink] New post   15 Mar 2016, 05:17
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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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sowragu
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Re: If m and n are the roots of the quadratic equation x2 - (2 root 5)x - [#permalink] New post   13 Mar 2016, 10:10
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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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stonecold
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Re: If m and n are the roots of the quadratic equation x2 - (2 root 5)x - [#permalink] New post   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] New post   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] New post   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

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

Answer: D
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Re: If m and n are the roots of the quadratic equation x2 - (2 root 5)x - [#permalink] New post   05 Aug 2022, 09:02
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