without solving the following quadratic equation, find the value of 'm' for which the given equation has real roots and equal roots : x^2 +2(m-1)x + (m+5)=0
Answers
Answered by
103
We know the condition
B^2-4ac>0
[2(m-1)]^2>4*(m+5)
4(m^2-2m+1)>4(m+5)
m^2-3m-4=0
m (-1,4)
B^2-4ac>0
[2(m-1)]^2>4*(m+5)
4(m^2-2m+1)>4(m+5)
m^2-3m-4=0
m (-1,4)
07161020:
its a too long ques
where
i can see it in my profile asked questions
https://brainly.in/question/1665619
I told u late that
I am so bored of doing my maths work lol
try my ques
Subject??
maths
Really i am sorry but i am fed up of maths
Answered by
157
Hey there,
That's a good question. So let's start from the very beginning!!!
EQUATION-x²+x(2m-2)+(m+5)
If the roots are to be real and equal then the value of discriminant(D) is to be 0.
D=0
(2m-2)²-4(m+5)=0
4m²-12m-16=0
Taking 4 as common, we get-
m²-3m-4=0
m²+4m-m-4=0
m(m+4)-1(m+4)=0
(m-1)(m+4)=0
Therefore m=1 or m=-4
PLEASE MARK AS BRAINLIEST IF HELPFUL!!!
Regards
07161020
Ace
That's a good question. So let's start from the very beginning!!!
EQUATION-x²+x(2m-2)+(m+5)
If the roots are to be real and equal then the value of discriminant(D) is to be 0.
D=0
(2m-2)²-4(m+5)=0
4m²-12m-16=0
Taking 4 as common, we get-
m²-3m-4=0
m²+4m-m-4=0
m(m+4)-1(m+4)=0
(m-1)(m+4)=0
Therefore m=1 or m=-4
PLEASE MARK AS BRAINLIEST IF HELPFUL!!!
Regards
07161020
Ace
thansk for marking brainliest
It was a great help for me
pleasure was mine
Similar questions