Question

If both the roots of the equation x² - (m+3)x-2m = 0 are positive and distinct, where m is a real number, then the correct option is​

Answer 1

Answer:

m ∈(-3 , -7+2$\sqrt{10}$)

Step-by-step explanation:

Given,

Both the roots of the equation x² - (m+3)x-2m = 0 are positive and distinct

To find,

The value of 'm'

Recall the concepts,

1. If the roots of the quadratic equation ax² +bx +c = 0 are real and distinct, then the discriminant D is positive. That is D = b² - 4ac >0
2. Also, the roots are positive if $\frac{-b}{2a} > 0\ and\ c > 0$

Solution:

Given equations is x² - (m+3)x-2m = 0

Comparing this equation with ax² +bx +c = 0

a = 1, b = -(m+3) and c = -2m

Then, b² - 4ac = (m+3)² - 4×1×-2m

= m² +6m +9 +8m

= m² +14m +9

b² - 4ac >0 ⇒ m² +14m +9 >0

By using the quadratic formula we get the roots of the equation m² +14m +9 as (-7-2$\sqrt{10}$) and (-7+2$\sqrt{10}$)

so,

m² +14m +9>0

(m+7+2$\sqrt{10}$)(m+7-2$\sqrt{10}$)>0

so,     m<(-7-2$\sqrt{10}$), m>(-7+2$\sqrt{10}$)--------------------(1)

The Roots are positive if $\frac{-b}{2a} > 0\ and\ c > 0$

$\frac{-b}{2a} > 0$

$\frac{m+3}{2}$>0

m+3>0

m>-3---------------------------------------------------------(2)

c>0

-2m>0

-m>0

m<0---------------------------------------------------------(3)

Therefore

Union of equations (1),(2), and (3) gives,

m belongs to the interval (-3, -7+2$\sqrt{10}$)

Hence,

m ∈(-3, -7+2$\sqrt{10}$)

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Answer 2

Answer:

Hence the value of m is $m$∈$(-3, -7+2\sqrt{10} )$

Step-by-step explanation:

• The equaton given to us is
• $X^2 - (m+3)X -2m =0$
• Now for the roots of the equation to be real, posetive and distinct
  $D > 0\\D=b^2-4*a*c\\b^2-4*a*c > 0$
• Now the value of
  $b=m+3\\a=1\\c=-2m$
• After substituting the value we get
  $(m+3)^2 +8m > 0\\m^2+14m+9 > 0$
• From here we get the value of m as
  $m < (-7-2), m > (-7+2)$
• but for the roots to be posetive
  $\frac{-b}{2a} > 0\\m > -3$   and  $m < 0$
• Hence the value of m is
  $m$∈$(-3, -7+2\sqrt{10} )$
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