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To have real solution, b^2 - 4ac >=0.(You know this)
To have two rational solutions, (b^2 - 4ac) should be a perfect square.
k^2 - 4×2×(-4) = k^2 + 32
For K=1, k^2+32 = 33 which is not a perfect square
For K=2, k^2+32 = 36 which is a perfect square. (One solution)
For K=-2, k^2+32 = 36 which is a perfect square. (Two solution)
For K=7, k^2+32 = 81 which is a perfect square (three solutions)
For K=-7, k^2+32 = 81 which is a perfect square (four solutions)
Total four solutions.
To have two rational solutions, (b^2 - 4ac) should be a perfect square.
k^2 - 4×2×(-4) = k^2 + 32
For K=1, k^2+32 = 33 which is not a perfect square
For K=2, k^2+32 = 36 which is a perfect square. (One solution)
For K=-2, k^2+32 = 36 which is a perfect square. (Two solution)
For K=7, k^2+32 = 81 which is a perfect square (three solutions)
For K=-7, k^2+32 = 81 which is a perfect square (four solutions)
Total four solutions.
billionaire:
y u took only 1,2,-2,7,-7 for value of k?
put values of k as 1.,2,3,4,5.... you will see that only 2 and 7 satisfy. since there is k^2, -2 and -7 will also satisfy
any more doubt?
ok great
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