Problem 8 p^2 - 30p + c Step One Take 1/2 of - 30 1/2 * -30 = - 15
Step 2 Square -15 (-15)^2 = 225 c = 225
Problem Nine [tex]\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a}
[/tex] a = 1 b = 4 c = -15 [tex]\text{x = }\dfrac{ -4 \pm \sqrt{4^{2} - 4*1*(-15) } }{2*1} [/tex] [tex]\text{x = }\dfrac{ -4 \pm \sqrt{\text{16} + \text{60} } }{2} [/tex]
x = [-4 +/- sqrt(76)] / 2 x = [-4 +/- 2*sqrt19]/2 x = [-4/2 +/- 2/2 sqrt[19] x = - 2 +/- sqrt(19)
x1 = - 2 + sqrt(19) x2 = -2 - sqrt(19)
These two can be broken down more by finding the square root. I will leave them the way they are. It's just a calculator question if you want it to go into decimal form.
Problem Ten
a = 1 b = 4 c = -32
The discriminate is sqrt(b^2 - 4ac) D = sqrt(b^2 - 4ac) D = sqrt(4^2 - 4(1)(-32) D = sqrt(16 - - 128) D = sqrt(16 + 128) D = sqrt(144) D = +/- 12
Since D can equal + or minus 12 there must be 2 possible (and different) roots. As a matter of fact, this quadratic can be factored. (x + 8)(x - 4) = y But that' s not what you were asked for. The discriminate is > 0 so the roots are going to be real. Answer; The discriminate is > 0 so there will be 2 real different roots.
