Question

The roots of the quadratic equation
2x^2+11x-40=0 are 2e and (f-2).Calculate the possible values of e and f. @ParthKohli

Answer 1

Again, you can use Vieta's Formulas if you want to.

Answer 2

Or use the quadratic formula.

Answer 3

I'd prefer the latter for this one.

Answer 4

sum of roots =-11/2
so 2e+f-2=-11/2------------(i)
4e+2f-4=-11
4e+2f=-7
f=(-7-4e)/2
now product of roots=-40/2
2e.(f-2)=-20
put the value from eq (i) of f in this now it become quadratic in f solve for f then put value of f in (i) get value of e.

Answer 5

I would solve the roots instead hmm

Answer 6

Yes.

Answer 7

By the quadratic formula.

Answer 8

looks it can be factored by grouping

Answer 9

For some reason, it's hard to factorise when you've got multiples of 10. I can only think of the factors that are multiples of 10.

Answer 10

I can confirm that my brain is autistic when it comes to factoring. The \(n^4 +4 \), uh...

Answer 11

Haha sometimes it requires irritating prime factorization and checking all combinations one by one i think 40*2 = 10*4*2 = 5*2^4

Answer 12

That's what I later did. But I can't ever think of it when a number is in a raw form.

Answer 13

same wid me too lol i end up with listing all divisors and selecting the right ones by checking each divisor

Answer 14

@MARC_ are you following us

Answer 15

Thank you @ParthKohli @rational @Er.Mohd.AMIR

Answer 16

2x^2+11x-40=0

2x^2 + 16x - 5x - 40 = 0

2x(x+8) - 5(x+8) = 0

(x+8)(2x-5) = 0

x = -8, 5/2

Answer 17

so the possible values of \(e\) and \(f\) can be obtained by :

\(-8 = 2e\) AND \(5/2 = f-2\)

or

\(-8 = f-2\) AND \(5/2 = 2e\)

Answer 18

The given answer in the book is \[e=\frac{ 5 }{ 4 }\]and\[f=\frac{ 9 }{ 2 }\]

Answer 19

Hmm we're getting :

\(\large e = -4,~~\frac{5}{4}\)

and

\(\large f = -6,~~\frac{9}{2}\)

Answer 20

oh i forget to type e=-4 and f=-6

Answer 21

Thnx @rational :)

Answer 22

np :) it will be a good exercise to solve this problem below methods and see each method gives the same answer :

1) Vieta's formulas - as started by Er.Mohd.AMIR
2) solving roots first and comparing - we did this
3) do you remember any other ways ?

Answer 23

can we use completing the square to solve this question @rational ?

Answer 24

yes we can solve the roots using below methods

1) factor by grouping
2) completing the square
3) quadratic formula

Answer 25

they all fall under second method though
```
1) Vieta's formulas - as started by Er.Mohd.AMIR
2) solving roots first and comparing - we did this
a) factor by grouping
b) completing the square
c) quadratic formula
3) do you remember any other ways ?
```

Answer 26

oh ok
Thnx @rational :)