Question

Find the value of k in (2k+4)^2 -(k-5)^2=26k

Answer 1

$\bold{Answer:}$

$k = \pm\sqrt{3}$

$\bold{Step} - \bold{by} - \bold{step\ explanation:}$

$(2k+4)^2-(k-5)^2=26k \\ \\ ((2k+4)+(k-5))((2k+4)-(k-5))=26k \ \ \ \ \ [a^2-b^2=(a+b)(a-b)] \\ \\ (2k+4+k-5)(2k+4-k+5)=26k \\ \\ (3k-1)(k+9)=26k \\ \\ 3k^2+26k-9=26k \\ \\ 3k^2-9=0 \\ \\ 3k^2=9 \\ \\ k^2=\frac{9}{3} = 3 \\ \\ k = \bold{\pm\sqrt{3}}$

$[ At where the identity \ a^2-b^2=(a+b)(a-b)\ is used, \\ \\ take \ a=2k+4\ \ \&\ \ b=k-5]$

$Hope this may be helpful. \\ \\ Thank you. Have a nice day. :-) \\ \\ \\ \\ \\ \#adithyasajeevan$

Answer 2

Answer:

k = \pm\sqrt{3}k=±

3

\begin{lgathered}(2k+4)^2-(k-5)^2=26k \\ \\ ((2k+4)+(k-5))((2k+4)-(k-5))=26k \ \ \ \ \ [a^2-b^2=(a+b)(a-b)] \\ \\ (2k+4+k-5)(2k+4-k+5)=26k \\ \\ (3k-1)(k+9)=26k \\ \\ 3k^2+26k-9=26k \\ \\ 3k^2-9=0 \\ \\ 3k^2=9 \\ \\ k^2=\frac{9}{3} = 3 \\ \\ k = \bold{\pm\sqrt{3}}\end{lgathered}

(2k+4)

2

−(k−5)

2

=26k

((2k+4)+(k−5))((2k+4)−(k−5))=26k [a

2

−b

2

=(a+b)(a−b)]

(2k+4+k−5)(2k+4−k+5)=26k

(3k−1)(k+9)=26k

3k

2

+26k−9=26k

3k

2

−9=0

3k

2

=9

k

2

=

3

9

=3

k=±

3