Question

if the roots of the equation x^2-15-m(2x-8)=0 are real and distinct then m is the element of

Answer 1

Step-by-step explanation:

x² -15 -m(2x-8)=0

=>x² -15 -2mx+8m=0

=>x² -2mx +(8m-15)=0

Comparing with the form ax²+bx+c= 0

we have

a= 1

b= -2m

c= 8m-15

for real distinct roots we have,

b² - 4ac ≥ 0

=>(-2m)² - 4(1)(8m-15) ≥ 0

=> 4m² -32m + 60 ≥ 0

=> m² - 8m + 15 ≥ 0

=> m² - 3m -5m +15 ≥ 0

=> m(m-3) -5(m-3) ≥ 0

=> (m-5) (m-3) ≥ 0

m= 5; m= 3

Case 1: take m=5

x² -15 -5(2x-8)=0

=>x² -15 -10x + 40=0

=>x² -10x + 25=0

=>x² -5x -5x + 25=0

=>x(x-5) -5(x-5)=0

=> (x-5)(x-5)=0

x = 5; x= 5

So, the equal roots are 5,5

Case 2: take m=3

x² -15 -3(2x-8)=0

=>x² -15 -6x + 24=0

=>x² -6x + 9=0

=>x² -3x -3x + 9=0

=>x(x-3) -3(x-3)=0

=> (x-3)(x-3)=0

x = 3; x= 3

So, the equal roots are 3,3

Therefore,the possible values of m= 3 and 5