Question

the set of all real values of a for which the equation
-(ax - 2) = 2x^2+ax+4.
please do it .....
I need it urgently
I'll mark u as BRAINLIEST.......​

Answer 1

Case 1, ax−2≥0.

ax−2=2x2+ax+4

0=2x2+6=2(x2+3)

x=±−3−−−√=±i3–√ 3

Imaginary roots — we can skip this case.

Case 2, ax−2<0

−(ax−2)=2x2+ax+4

0=2x2+2ax+2=x2+ax+1  

x=12(−a±a2−4−−−−−√)

For a positive root, we need the plus sign and also

−a+a2−4−−−−−√>0

a2–4−−−−√>a

(EDIT to fix a second mistake.) This is only going to work when a<0. We require the discriminant to be non-negative.

a2−4≥0

a2≥4

At a=−2 we should get exactly one positive root. Pls mark brainliest

Hope helps

Took lots of time for solving

Answer 2

Step-by-step explanation:

⭕Case 1

, ax−2≥0.

ax−2=2x2+ax+4

0=2x2+6=2(x2+3)

x=±−3−−−√=±i3–√

3

Imaginary roots — we can skip this case.

⭕Case 2

, ax−2<0

−(ax−2)=2x2+ax+4

0=2x2+2ax+2=x2+ax+1  

x=12(−a±a2−4−−−−−√)

For a positive root, we need the plus sign and also

−a+a2−4−−−−−√>0

a2–4−−−−√>a

a2−4≥0

a2≥4

At a=−2 we should get exactly one positive root.