Question

For what value of k does (k-12)x square - 2 (k-12)x+2=0 have equal roots?(k is not=12

Answer 1

Answer:

10

Step-by-step explanation:

Equation: (k-12)x²-2(k-12)x+2

So, a =k-12, b =-2(k- 12) and c =2

For equal roots, b²-4ac = 0

= (-2(k-12))-4(k-12)(2) = 0

= (-2k+24)-4(2k-24) = 0

= -2k+24-8k+72 = 0

= -10k+100 = 0

= k = -100/-10

= k = 10

So, k = 10 foes the equation have equal roots.

Answer 2

Answer:

Either k=4 or k=-30

Step-by-step explanation:

It is given that the equation has equal roots

• i.e b^2-4ac=0

the given equation a is equal to K - 12

b=-2(k-12) and c= 2

• (-2(k-12))^2-4(k-12)(2)=0
• (-2k+24)^2-8k-96=0
• 4k^2+576-2(2k)(24)-8k-96=0
• 4k^2+576-96k-8k-96=0
• 4k^2-104k+480=0
• k^2-26k+120=0. Dividing the equation by 4
• k^2-30k+4k+120=0
• k(-k-30)-4(-k-30)=0
• (k-4)(-k-30)=0
• Either k-4=0 or -k-30=0
• Therefore, either k=4 or k=-30