Question

if the quardratic equation (k-3)x2 + 4(k-3)x +4=0 has real and equal roots find the value of k
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Answer 1

For equal roots ,

${b}^{2} - 4ac = 0$

• a = k-3
• b = 4(k-3)
• c = 4

${4}^{2} ( {k - 3)}^{2} - 4(k - 3)(4) = 0 \\ \\ 16( {k}^{2} - 6k + 9) - 16k + 48 = 0 \\ .....dividing \: both \: sies \: by \: 16... \\ \\ {k}^{2} - 6k + 9 - k + 3 = 0 \\ \\ {k}^{2} - 7k + 12 = 0 \\ \\ {k}^{2} - 4k - 3k + 12 = 0 \\ \\ k(k - 4) - 3(k - 4) = 0 \\ \\ (k - 3)(k - 4) = 0 \\ \\ k = 3...........k = 4$

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Answer 2

Answer:

4

Step-by-step explanation:

a=k-3

b=4(k-3)

c=4

b²-4ac=0

b²=4ac

16(k-3)(k-3)=4(k-3)(4)

16(k-3)(k-3)=16(k-3)

16(k-3) get cancelled on both sides

k-3=1

k=4

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