Answer :
[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]
[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)
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)
