Question

If the roots of the equation x²+(m-3)x+(m+1)= 0 are equal, then the value of
m² is​

Answer 1

GiveN :-

• The roots of the equation are equal.
• The equⁿ is x²+(m-3)x + (m+1)=0.

To FinD :-

• The value of m.

AnsweR :-

Given quadratic equation is x²+(m-3)x + (m+1)=0.. So , we know that if the roots of the Quadratic Equation are equal , then the value of Discriminant is 0 . For a quadratic equation in standard form ax² + bx + c , Discriminant is b² - 4ac and here in this case , Discriminant=0.

Here , a table for Nature of roots :-

$\begin{tabular}{|c|c|}\cline{1-2} \sf Discriminant>0 & \sf Roots are equal \\\cline{1-2} \sf Discriminant =0 & \sf Roots are equal\\\cline{1-2} \sf Discriminant < \: 0 &\sf Roots are complex \\\cline{1-2}\end{tabular}$

Hence , now we know that for equal roots D = 0, With respect to Standard form ax² + bx + c = 0 , here ;

• a = 1
• b = ( m - 3 )
• c = ( m + 1 )

⇒ b² - 4ac = 0

⇒ ( m - 3 )² - 4*1*(m+1) = 0

⇒ m² + 9 -6m - 4m -4 = 0

⇒ m² - 10m +5 = 0

⇒ m = 10 ± √[ 100 - 20 ] / 2 ( By using Quadratic formula)

⇒ m = 10 ± √ 80 / 2

⇒ m = 10 ± √ [2⁴ * 5] /2

⇒ m = 10 ± 4√5 /2

⇒ m = 5 ± 2√5

⇒ m = 5 + 2√5 , 5 - 2√5

$\Large\boxed{\pink{\sf \red{\bigstar}\:Value\:of\:m\:=\:5\pm 2\sqrt5}}$