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how should i start

For a root with multiplicity two you need to need to have the discriminant equal to 0.

D=b^2-4ac

but what is b and a and c the equation is a bit confusing

Since we want root with multiplicity 2 :)
So \[B^2-4AC=0\]

\[\text{ you have } x^2+(7+a)x+(7a+1)=0\]

A is the number in front of x^2
B is the number in front of x
C is the leftover part

\[x^2+(7+a)x+7a+1=0\]

A=1
B=(7+a)
c=(7a+1)

Yep! :)

\[(7+a)^2-4(1)(7a+1)=0\]

That is B^2-4AC=0
You think you can solve that reply above for a?

ok wait

49+14a+a^2-14a+4=0

I think you mean -28a since -4(7)=-28 not -14

yeah -28a

a^2-14a+53=0

is it right

well .... I think your 53 is a little off.
you have 49-4
you forgot to distribute that - earlier

\[49+14a+a^2-4(7a+1)=0\]
\[49+14a+a^2-28a-4=0\]
Now try combining like terms.

a^2-14a+45=0

yep
do you know how to factor?

yep but i am solving this with quadratic is it ok?

The quadratic formula?
That is fine. :)

yes

I solved it by factors..

thnx @myininaya U are great. Nice help

thanks

got 2 values 9 and 5 of a

Great! You go guys! :)

The discriminant is \[B^2-4AC \]

How did it came.

@myininaya how to solve the equation to get the discriminant..

So to slove the equation should i find a b c

*solve