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# Help me!!

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Hello! I am having a difficult time with these questions. I don't understand them very well. Please help! I apologize for taking up your time. Please show the work because I have other questions similar to these and I would like to work on the other ones myself.  Thank you!

Sep 12, 2018

#1
+101161
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DC  = BC  ...  thus... they will  both be  =  DB  / √2    [ the diagonal of a square is always √2 * side length ....so...the side length  =  diagonal / √2  ]

So

If  DB  = 4  then   DC = BC  = 4 / √2  = [ rationalize the denominator] = 4√2 / 2 = 2√2...so...false

If DB  = 14  then DC = BC  =  14 / √2   =  [rationalize the denominator]  =  14 * √2  / [ √2 * √2 ] =

14√2 / 2   =  7√2....so......this is true

If DB  = 4√2  then DC = BC  =  4√2 / √2  =  4....so....true

If DB  = √10  then DC = BC  = √10 /√2   =  √5   ....so...true

If DB = 15 then  DC = BC  =  15 / √2  =  15√2 / 2.....so true

If DB  = √5 then  DC = BC  = √5 / √2   =   √5 * √2  / 2  =  √10 / 2...so....false

Sep 12, 2018
#2
+101161
+1

Second one :

Put these  in the form   ax^2  + bx + c  = 0

The discriminant  is given by   [ b^2 - 4ac ]...if this is > 0, we will have two real roots

3x^2 - 4x + 7  =  0

discriminant  =  (-4)^2 - 4(3)(7)  =  16 - 84  ....not > 0....so...no two real roots

1x^2 - 2x + 8  = 0

discriminant  =  (-2)^2 - 4(1)(8)  =  4 - 32  ....not > 0...so....no two real roots

5x^2 + 6  =  0

discrimant  =  (0)^2  - 4(5)(6)  = - 120  ....not > 0....no real roots

4x^2  - 9  = 0

discriminant  =  (0)^2 - 4(4)(-9)  = 36 > 0  ....  so....two real roots

2x^2 + 3x  - 4  = 0

discriminant = (3)^2  - 4(2)(-4)  =  9 + 32  > 0...so...two real roots

1x^2 - 5x - 3  = 0

discriminant =  (-5)^2 - 4(1)(-3)  = 25 + 12  > 0....so.....two real roots

Sep 12, 2018