Question

plz solve 1 and 2
step by step

Answer 1

(1)
b²- 4ac = 0
a = k-12
b = 2(k-12)  => b = 2k - 24
c = 2


= (2k-24)² - 4×(k-12)×2 = 0
=4k²-96k+576 - 8 × (k-12) = 0
=4k²-96k+576 - 8k +96 = 0
=4k² - 104k +672 = 0
divide by 4 throughout on both sides
= k² - 26k + 168 = 0
=k² - 12k - 14k +168 = 0
=k(k-12) -14( k-12) = 0
=>(k-12)(k-14)
therefore k = 12 or 14 for the first one

(2)
a=k²
b=2(k-1) => 2k-2
c=4

b²- 4ac = 0
=(2k-2)² - 4 × k² × 4 = 0
=4k² - 8k + 4 -16k² = 0
= -12k² - 8k +4 = 0
=12k² + 8k - 4 =0
divide by 4 throughout
= 3k² + 2k - 1=0
=3k² + 3k - 1k -1 =0
= 3k(k+1) -1(k+1) = 0
=(3k-1)(k+1)
=> k= (1/3) or (-1)

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