if a and b are zeros of polynomial Kx²+4x+4 find the value of k such that (a+b)²+2ab=24
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a+b=-4/k and a×b=4/k
now, (a+b)²+2ab=24
or, (-4/k)²+2×4/k=24
or, 16/k²+8/k=24
or, (16+8k)/k²=24
or, 16+8k=24k²
or, 24k²-8k-16=0
or, 8(3k²-k-2)=0
or, 3k²-k-2=0
or, 3k²-3k+2k-2=0
or, 3k(k-1)+2(k-1)=0
or, (k-1)(3k+2)=0
either, k-1=0; or, k=1
or, 3k+2=0; or, k=-2/3
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