Question 1 Andregradslikningen ax2+bx+c=0 har løsningene x=−b±√b2−4ac2a, når b2−4ac≥0. Dersom andregradsuttrykket ax2+bx+c har nullpunktene x=x1 og x=x2 kan det faktoriseres til a(x−x1)(x−x2). a) Løs andregradslikningen x2−3x−28=0. Skriv svaret nedenfor som x1 og x2, da x1<x2 x1= Expected answer: (NB: x1<x2) x2= Expected answer: Save answer Score: 0/1 Feedback for a). Show feedback.The feedback has changed.This feedback is based on your last submitted answer. Save your changed answer to get updated feedback. Or, you could: ⤺ Go back to the previous part There's nothing more to do from here.b) Faktoriser andregradsuttrykket x2−3x−28. x2−3x−28=interpreted asExpected answer:interpreted as(x+4)(x−7) Save answer Score: 0/1 Feedback for b). Show feedback.The feedback has changed.This feedback is based on your last submitted answer. Save your changed answer to get updated feedback. Or, you could: ⤺ Go back to the previous part There's nothing more to do from here.c) Løs ulikheten x2−3x−28>0. Svar: x> Expected answer: Gap 2 Answer for part Gap 2 Ogeller Expected answer: Ogeller x< Expected answer: Save answer Score: 0/3 Feedback for c). Show feedback.The feedback has changed.This feedback is based on your last submitted answer. Save your changed answer to get updated feedback. Or, you could: ⤺ Go back to the previous part There's nothing more to do from here. Advice Score: 0/5 Total 0/5 Move to the next questionTry another question like this oneReveal answers