Question

find the value of k so that the quadratic equation ky(y-7)+49=0 has two equal roots

Answer 1

HEYA!!
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Ky ( y-7 ) + 49

$ky {}^{2} - 7ky + 49 \\ \\ for \: the \: equation \: to \: have \: equal \: roots \: \\ \\ b {}^{2} - 4ac = 0 \\ \\ ( - 7k) {}^{2} - 4 \times k \times 49 \\ \\ 49k {}^{2} - 196k = 0 \\ \\ 7k(7k - 28) = 0 \\ \\ 7k - 28 = 0 \\ \\ 7(k - 4) = 0 \\ \\ k = > 4$
Hence the value of K = 4

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Answer 2

ky(y-7)+49=0. We know, when there are two equal roots, b^2-4ac=0.

(here b^2 denote b square)

so, ky(y-7)+49=0

ky^2-7y+49=0. Now, b=-7k, a=k and c=49.

After putting the values in b^2-4ac=0,

we get, 49k^2-196k= 0

7k(7k-28)=o

7k-28=o

k-4=o

so, k=4

Verification, now, b=-28, a=4 and c=49.

After putting the values in b^2-4ac=0,

we get, -28^2-4x4x49

784-784=0

I hope it helps you.

Plz pardon, if something went wrong.