How to stop quadratic formula calculator from giving inverse outputs?

Hi, I'm doing a worksheet based around matlab which involved solving quadratic formulas. I probably overcomplicated it and now made a calculator that keeps giving the inverse of my expected answer, could anyone help please?

for k = 1 : 7
disp ( ' For the equation ax^2 + bx + c ' )
a = input ( ' Enter a; ' );
b = input ( ' Enter b; ' );
c = input ( ' Enter c; ' );
D = ( b^2 ) - ( 4 * a * c );

if D < 0
fprintf ( ' \n The equation has no real roots . \n\n ')

elseif D == 0
root = -b / ( 2 * a );
fprintf ( ' \nThe equation has one root, \n ' )
fprintf ( ' %.7f\n\n ' , root )

else 
r1 = ( - b + sqrt ( D ) ) / ( 2 * a ) ;
r2 = ( - b - sqrt ( D ) ) / ( 2 * a ) ;
fprintf ( '\n The equation has two roots, \n ' ) 
fprintf ( ' %.7f and %.7f\n\n ' , r1 ,r2 ) 

end 

end

For the equation ax^2 + bx + c 
 Enter a; 
1
 Enter b; 
9
 Enter c; 
14

 The equation has two roots, 
  -2.0000000 and -7.0000000

  For the equation ax^2 + bx + c 
 Enter a; 
1
 Enter b; 
-3
 Enter c; 
-18

 The equation has two roots, 
  6.0000000 and -3.0000000

[Dyuman Joshi]

What are the expected answers for the examples you have used?

And how are the output values (from your code) inverse of them?

[Alice]

on the first one it was this that was inversed, as i expected two positive roots. im an absolute beginner with this sorry

Enter a;
1
Enter b;
9
Enter c;
14

The equation has two roots, 
  -2.0000000 and -7.0000000

[Alice]

nevermind, just realised my sleep deprived misconception, sorry !

Answer 1

[John D'Errico]

The roots of the quadratic you supply are indeed 6 and -3.

roots([1 -3 -18])

ans = 2x1

6.0000 -3.0000

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

So your code worked correctly.

If you expected something else, then it is your expectations that are wrong, not your code.

Anyway, perhaps I should point out that a cute feature of polynomial equations in that if you reverse the sequence of the cofficients, thus as here...

roots([-18 -3 1])

ans = 2x1

-0.3333 0.1667

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

then the computed roots will be the inverses of the true roots. And that is what I think has happened to you in some way. This would be true for any order polynomial, in fact.

Perhaps that is what you did at some point in your code, or maybe in your spreadsheet, you provided the coefficients in the reverse order. Can we know what you actually did wrong, since your code works correctly? Only da shadow know, but I'm going to bet, that IF you have gotten the inverse of the expected roots, then somewhere, somehow, you flipped the sequence of the coefficients. I'm not da shadow though.

[Alice]

hi , it was more the (x^2 + 14x+ 9) coming out as -2.0000... and -7.0000.... that i was worried about, as most equations that i expected to have two positive roots like that one came out with negatives

[Alice]

nevermind, just realised my sleep deprived misconception, sorry !

[John D'Errico]

Lol. An intersting fact about an equation like

x^2 + 14*x + 9

is if ALL of the coefficients are positive, then there can NEVER be any positive real roots. Any real roots must either be zero or negative, and zero only happens if there is a zero constant term.

(You can prove that easily enough if you think about it.) For a positive root to exist, at least SOME of the coefficients must be negative.