Question

Give that one root of the equation mx^2 +nx+9=0 is triple the other, show that n^2=48m

Answer 1

Answer

We know that while finding the root of a quadratic equation ax2+bx+c=0 by quadratic formula x=2a−b±b2−4ac, 

if b2−4ac>0, then the roots are real and distinct 

if b2−4ac=0, then the roots are real and equal and

if b2−4ac<0, then the roots are imaginary.

Here, the given quadratic equation x2−mx+9=0 is in the form ax2+bx+c=0 where a=1,b=−m and c=9.

  

(i) If the roots are equal then b2−4ac=0, therefore,

b2−4ac=0⇒(−m)2−(4×1×9)=0⇒m2−36=0⇒m2=36⇒m=±36⇒m=±6

(ii) If the roots are distinct then b2−4ac>0, therefore,

b2−4ac>0⇒(−m)2−(4×1×9)>0⇒m2−36>0⇒m2>36⇒m>±36⇒m>±6

(iii) If the roots are imaginary then b2−4ac<0, therefore,

b2−4ac<0⇒(−m)2−(4×1×9)<0⇒m2−36<0⇒m2<36⇒m<±36⇒m<±6  

(iii) If the roots are imaginary then b2−4ac<0, therefore,

b2−4ac<0⇒(−m)2−(4×1×9)<0⇒m2−36<0⇒m2<36⇒m<±36⇒m<±6  

HOPE IT HELPS YOU