If 2 roots of an equation x^2-3x-3=0 are m and n, what is the value of

Question

If 2 roots of an equation x^2-3x-3=0 are m and n, what is the value of m^2+n^2?

A. 10 
B. 12 
C. 15 
D. 24 
E. 25

* A solution will be posted in two days.

MathRevolution · Accepted Answer

==>From x^2-3x-3=(x-m)(x-n)=x^2-(m+n)x+mn, you get m+n=3 and mn=-3. Then you get m^2+n^2=(m+n)^2-2mn=3^2-2*(-3)=15. The answer is C.

Answer: C

acegmat123 · Answer

m = x + y

n= x - y

Where x= -b/2a

y= (sqrt(b^2 - 4ac))/2a

m^2+n^2 = 2(x^2 + y^2)

= 2(9/4 + 21/4)

=15

C

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Abhishek009 · Answer

With  acegmat123

That's the perfect and easiest way of solving this question
 
With (C) hence...

PS : Good question   MathRevolution  ...

AK125 · Answer

Can someone post a step by step solution pls...Kudos for this

MathRevolution · Answer

Hi AK125,

x^2-3x-3=(x-m)(x-n)=x^2-(m+n)x+mn so, m-n=3 and mn=-3

Thank you.

hatemnag · Answer

It is not clear enough. I miss some step above. please any help ?

gillyelephant · Answer

I don't understand when you solve for (m^2 + n^2) and use (m+n)^2 - 2mn in its place

Posted from my mobile device

AK125 · Answer

We do know m+n which equals 3. And we know mn which is -3.

If we square (m+n) -->

( m+n )^2 = m^2 + 2 mn  + n^2

Plug in what we do know (green):

(3)^2 = m^2 + 2(-3) + n^2
--> 9 = m^2 - 6 + n^2
-->  15 = m^2 + n^2

Hope its clear!

spoorti · Answer

Sum of the roots = 3 (m+n)
Product of the roots = -3 (mn)
m^2 + n^2 = (m+n)^2 -2mn = 9+6 = 15

Posted from my mobile device

ritz07 · Answer

Its easier to use the formula for roots.

x has two solutions, x =   / 2a

hence, m =  /2(1)
 =   / 2
Similarly, n =   /2
so, m^2 + n^2 =  /4 = 60/4 = 15

bumpbot · Answer

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If 2 roots of an equation x^2-3x-3=0 are m and n, what is the value of

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If 2 roots of an equation x^2-3x-3=0 are m and n, what is the value of

Prep Toolkit

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