hey_thr67 wrote:Find all the values of 'a', so that 6 lies between the roots of the equation x^2 + 2(a-3)x + 9 =0

A: a< −3/4
B: a> 3/4
C: a<0 or a>6
D: a>6
E: a < −1/4

Good question. It tests our fundamentals of quadratic functions.

Note: Quadratic Functions are parabolic in nature. If a > 0, then the parabola points downwards and if the roots are real and distinct then the parabola intersects the X-axis at two points.

If
f(x) = ax^2 + bx + c, and if a > 0 and real roots exist, then
for each 'k' between the roots,
f(k) < 0



In this problem ,
f(x) = x^2 + 2(a - 3)x + 9 = 0
(a = 1 > 0)
f(6) < 0 for '6' to lie between the roots.
i.e.
6^2 + 2(a - 3)*6 + 9 < 0
15 + 4(a - 3) < 0
4a - 12 + 15 < 0
a < - 3/4

[spoiler](A)[/spoiler] is the answer.

hey_thr67 wrote:Find all the values of 'a', so that 6 lies between the roots of the equation x^2 + 2(a-3)x + 9 =0

A: a< −3/4
B: a> 3/4
C: a<0 or a>6
D: a>6
E: a < −1/4

Thanks for this question....where did you get this question from ?

aneesh.kg wrote:

hey_thr67 wrote:Find all the values of 'a', so that 6 lies between the roots of the equation x^2 + 2(a-3)x + 9 =0

A: a< −3/4
B: a> 3/4
C: a<0 or a>6
D: a>6
E: a < −1/4

Good question. It tests our fundamentals of quadratic functions.

Note: Quadratic Functions are parabolic in nature. If a > 0, then the parabola points downwards and if the roots are real and distinct then the parabola intersects the X-axis at two points.

If
f(x) = ax^2 + bx + c, and if a > 0 and real roots exist, then
for each 'k' between the roots,
f(k) < 0



In this problem ,
f(x) = x^2 + 2(a - 3)x + 9 = 0
(a = 1 > 0)
f(6) < 0 for '6' to lie between the roots.
i.e.
6^2 + 2(a - 3)*6 + 9 < 0
15 + 4(a - 3) < 0
4a - 12 + 15 < 0
a < - 3/4

[spoiler](A)[/spoiler] is the answer.

Thanks a lot Aneesh..
Can you please shed light on other such conditions..
ax2+bx+c=0
a>0 parabola downwards
a<0 parabola upwards
if b2-4ac>=0 real solution will exist.

1947 wrote: Thanks a lot Aneesh..
Can you please shed light on other such conditions..
ax2+bx+c=0
a>0 parabola downwards
a<0 parabola upwards
if b2-4ac>=0 real solution will exist.

There you go:



I have discussed about the vertical/horizontal shifting of parabolas and their shapes here:
https://www.beatthegmat.com/learn-how-to ... tml#478024