Question

find the value of k if 0=(2K-8)^2-16(K-4)​

Answer 1

(2k-8)²-16(k-4) =0

4k²+64-32k-16k+64 =0

4k²-48k+128=0

k²-12k+32=0

k= 12±√(144-128) / 2

= 12±4 / 2

= 8,4

so k=8,4

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Answer 2

Step-by-step explanation:

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To Find -

Value of k

$(2k - 8) {}^{2} - 16(k - 4) = 0\\ \\ \leadsto \: (2k) {}^{2} + (8) {}^{2} - 2 \times 2k \times 8 - 16k + 64 = 0 \\ \\ \leadsto4k {}^{2} + 64 - 32k - 16k + 64 = 0 \\ \\ \leadsto 4k {}^{2} - 48k + 128 = 0 \\ \\ now \bold \:{ factorising \: this} \\ \\ \leadsto4k {}^{2} - 16k - 32k + 128 \\ (by \: middle \: term \: split)\\ \\ \leadsto4k(k - 4) - 32(k - 4) \\ \\ \leadsto (k - 4)(4k - 32) \\ \\ k - 4 = 0 \: \: \: and \: \: \: 4k - 32 = 0 \\ \\ k = 4 \: and \: k = \frac{32}{4} \\ \\ k = 4 \: \: \: and \: k \: = 8$