Solve the quadratic equation y=11-2x-2x^2 c)(i) x = -2.9 or 1.9 (ii) y+(3−2x−2x2)=11−2x−2x2 y=11−2x−2x2−3+2x+2x2 y=8...

Solve the quadratic equation
y=11-2x-2x^2

c)(i) x = -2.9 or 1.9

(ii) y+(3−2x−2x2)=11−2x−2x2
y=11−2x−2x2−3+2x+2x2
y=8
∴x=-1.8 or 0.8
(iii) y=11−2x−2x2
dydx=−2−4x
Gradient at x = 1 : −2−4(1)=−2−4=−6
Please my problem is(ii)
How did they get x=-1.8 and x=0.8?

To get notifications when anyone posts a new answer to this question,
Follow New Answers

Post an Answer

Please don't post or ask to join a "Group" or "Whatsapp Group" as a comment. It will be deleted. To join or start a group, please click here

JAMB 2020 CBT Mobile App - Download Now, It's Free!
JAMB 2020 CBT Software Agents - Click Here to Apply

Answers (1)

EmX
2 weeks ago
It doesn't make alot of sense to me either.

But what they did is quite clear
(ii) you'll be getting is y=8. With that value ignore everything in the bracket on the Left hand side.

What you'll have is

y=11-2x-2x².

With that you can substitute y=8 and solve.
Ask Your Own Question

Quick Questions

See More Mathematics Questions
 
JAMB 2020 CBT Software - Download Now, It's Free!
JAMB 2020 CBT Mobile App - Download Now, It's Free!
JAMB 2020 CBT Software Agents - Click Here to Apply