maths mc

If 2x^2 - qx + 8 has no root, then 2x^2 - qx + 8 > 0 (since a = 2 is positive)

discriminant < 0

q^2 - 64 < 0

(q - 8)(q + 8) < 0

-8 < q < 8

The answer is A

2x^2-qx+8>0

2x^2-qx+8 is always positive.

which means the graph 2x^2-qx+8 does not touch any point of x-axis .

2x^2-qx+8 has no real root(s)

Δ < 0

(-q)^2 - 4 (2)(8) < 0
q^2 - 64 < 0
(q+8)(q-8) < 0

.`. -8<q<8

Ans. is A

2012-04-13 19:59:45 補充：
Also , the cure y=2x^2-qx+8 is open upward.

2012-04-13 20:07:05 補充：
You can also use the method of completing the square to find the vertex of the curve.

2x^2-qx+8

= 2(x-q/4)^2-q^2+8

vertex= (q/4 , -q^2/8+8)

As 2x^2-qx+8>0,

y intercept of the vertex most >0

solving

-q^2/8+8 > 0

-8<8

2012-04-13 20:10:58 補充：
2x^2-qx+8>0

means the graph

y = 2x^2-qx+8

is always above y=0

2012-04-13 20:13:12 補充：
A.-8<8

2012-04-13 20:13:38 補充：
therefore no real roots

2012-04-13 20:15:43 補充：
2x^2-qx+8>0 cannot stand for the one of the roots of the equation<0 and the other root>0?

No, as explained above.