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If the equation x^2+(7+a)x+7a+1=0 has equal roots,find the value of a. . . . ANSWER = {9,5}

Mathematics
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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

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Other answers:

So what I'm saying is: \[Ax^2+Bx+C=0 => \text{ discriminant } = B^2-4AC\] And we want here for the discriminant=0
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! :)
If you guys didn't understand why I set the discriminant equal to 0, then I will try to explain as best as possible. If discriminant is equal to 0, then you will have on root (multiplicity 2) If discriminant is equal to negative number, you have two imaginary solutions If discriminant is equal to positive number, you have two real solutions.
The discriminant is \[B^2-4AC \]
How did it came.
@myininaya how to solve the equation to get the discriminant..
\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \text{ is called the quadratic formual } \] The discriminant is that thing under the radical It is what determines if you have one root (w/ multiplicity 2) , imaginary roots, or real roots.
So to slove the equation should i find a b c
*solve

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