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!
+128475 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
+128475 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
