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after simplification,we get the quadratic eqn

k^2-4k-65= 0

sum of all values of k will be equal to the sum of the roots of the above q.eqn = -b/a = 4

I will go with 4
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I tried solving this by quadratic equation formula and got 2 answers - 2+ or -\sqrt{65}.

What could be the other two possible values of k?
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5(k+2)^2 = 4k^2 + 24k + 85

=> 5(k^2 + 4k + 4) = 4k^2 + 24k + 85

=> 5k^2 + 20k + 20 = 4k^2 + 24k + 85

=> k^2 - 4k - 65 = 0

No need to solve this, we need sum of two values of k:

So sum of roots = -b/a = -(-4) = 4
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f(x)=5x^2
g(x)=x^2+12x+85
f(k+2)=5 (k+2)^2= 5(k^2+4k+4)=5k^2+20k+20

g(2k)= (2k)^2+ 12(2k)+85=4k^2+24k+85

5^k2+20k+20=4k^2+24k+85
k^2-4k-65=0

Using the quadratic equation (-b +/-\sqrt{b2-4ac}/2a we find the roots as follows

-(-4)+/-\sqrt{4^2 - 4(1)(-65)}/2(1)
So the first root is
4+\sqrt{16+260}/2 = 4+\sqrt{276}/2= 4+2\sqrt{69}/2
The second root is
4-\sqrt{16+260}/2 = 4-\sqrt{276}/2= 4-2\sqrt{69}/2

The sum of the roots is
4+2\sqrt{69}/2 + 4-2\sqrt{69}/2 = 8/2= 4
The answer should be 4 :-D
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sum of two roots of a quadratic equation is -b/a


now we have \(k^2-4k-65\)

=> - b/a = 4
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How do we get 4??

-b/a?? what is concept behind it??

k2-4k-65=0?? i cant solve after that..

plz do explain..Thanks in advance
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How do we get 4??

-b/a?? what is concept behind it??

k2-4k-65=0?? i cant solve after that..

plz do explain..Thanks in advance

It's called Viete's theorem, which states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):

\(x_1+x_2=\frac{-b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

Now, if we apply this to \(k^2-4k-65=0\), we'll get that the sum of the roots must be \(k_1+k_2=\frac{-(-4)}{1}=4\).

Hope it's clear.
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Thanks Alot BUNUEL..

i didnt see formula like this in my life :( new things very hard to digest :(..
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\(f(k+2) = 5(k+2)^2 = 5(k^2 + 4k + 4) = 5k^2 + 20k + 20\)
\(g(2k) = 4k^2 + 24k + 85\)

\(5k^2 + 20k + 20 = 4k^2 + 24k + 85\)
\(k^2 - 4k -65 = 0\)

Trick: Sum of all roots: k1 + k2 = -b/a

\(k1 + k2 = - \frac{-4}{1} = 4\)

Answer: 4
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Bunuel ,

Can we have more this kinda questions of functions ?
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Is this formular appliable even b^2 - 4c < 0? Bunuel ?
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Aalto700
Is this formular appliable even b^2 - 4c < 0? Bunuel ?
\(b^2 - 4ac\) is discriminant, if \(b^2 - 4ac < 0\), then the quadratic equation has no real roots. Vieta's formula is still valid, it will give you the sum / product of two complex roots.
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You are right! Thank you Krunaal!
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Aalto700
Is this formular appliable even b^2 - 4c < 0? Bunuel ?
\(b^2 - 4ac\) is discriminant, if \(b^2 - 4ac < 0\), then the quadratic equation has no real roots. Vieta's formula is still valid, it will give you the sum / product of two complex roots.
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