Answered

Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

Determine the exact value of k so that the quadratic function f(x) = x2 - kx + 5 has only one zero.

Sagot :

[tex]f(x)=ax^2+bx+c\\\\\Delta=b^2-4ac\\\\if\ \Delta < 0-no\ zeros\\if\ \Delta=0-one\ zero\\if\ \Delta > 0-two\ zeros[/tex]


[tex]f(x)=x^2-kx+5=0\\\\a=1;\ b=-k;\ c=5\\\\\Delta=(-k)^2-4\cdot1\cdot5=k^2-20\\\\one\ zero\ if\ \Delta=0\\\\therefore\\k^2-20=0\ \ \ \ |add\ 20\ to\ both\ sides\\\\k^2=20\\\\k=\pm\sqrt{20}\\\\k=\pm\sqrt{4\cdot5}\\\\\boxed{k=-2\sqrt5\ or\ k=2\sqrt5}[/tex]
I like this question. When we factorise this question the brackets have to be identical.
5 has to be square rooted to become √5. From, FOIL we know that the last digit is times by the other last digit to find the 5, as our brackets are identical this number is the same. The square root of 5.
This number is doubled in identical brackets to find the middle number. so it is 2√5. As there is a minus number there the brackets are: (x-√5)(x-√5). Multiplying this out gives us: x²-2√5 x+5. k=2√5 (or -2√5, depending on if the minus is counted or not)