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Find all values of k that ensure that the given equation has exactly one solution. 4x2 + kx + 49 = 0 What is: k = ____ (smaller value) k = ____ (larger value)
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k^2 -4(4)(49) =0
solve , you will get one positive and one negative , thus the largest and smallest
k^2 - (4^2 )(7^2) = 0 k ^2 - 28^2 =0 k=+-28
being a pro without a calculator :D
make sure \[b^2-4ac=0\] with \[a=4,b=k,c=49\]
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\[k^2-4\times 4 \times 49=0\] \[k^2-4^2\times 7^2=0\] \[k^2=4^2\times 7^2\] \[k=\pm 4\times 7\] \[k= \pm 28\]
what elecengineer wrote!
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