If the equation
has real roots then find the value of k.
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Answered by
4
Hey friend !
Your answer is given below.
![equation = {kx}^{2} - 5x + k = 0 \\ \\ a =k \\ b = - 5 \\ c = k \\ \\ farmula = {b}^{2} - 4ac \\ \\ { - 5}^{2} - 4 \times k \times k \\ 25 - {4k}^{2} =0 \\ {4k}^{2} = 25 \\ k = \sqrt[ + - ]{ \frac{25}{4} } \\ k = \frac{5}{2}](https://tex.z-dn.net/?f=equation+%3D+%7Bkx%7D%5E%7B2%7D+-+5x+%2B+k+%3D+0+%5C%5C+%5C%5C+a+%3Dk+%5C%5C+b+%3D+-+5+%5C%5C+c+%3D+k+%5C%5C+%5C%5C+farmula+%3D+%7Bb%7D%5E%7B2%7D+-+4ac+%5C%5C+%5C%5C+%7B+-+5%7D%5E%7B2%7D+-+4+%5Ctimes+k+%5Ctimes+k+%5C%5C+25+-+%7B4k%7D%5E%7B2%7D+%3D0+%5C%5C+%7B4k%7D%5E%7B2%7D+%3D+25+%5C%5C+k+%3D+%5Csqrt%5B+%2B+-+%5D%7B+%5Cfrac%7B25%7D%7B4%7D+%7D+%5C%5C+k+%3D+%5Cfrac%7B5%7D%7B2%7D+ "equation = {kx}^{2} - 5x + k = 0 \\ \\ a =k \\ b = - 5 \\ c = k \\ \\ farmula = {b}^{2} - 4ac \\ \\ { - 5}^{2} - 4 \times k \times k \\ 25 - {4k}^{2} =0 \\ {4k}^{2} = 25 \\ k = \sqrt[ + - ]{ \frac{25}{4} } \\ k = \frac{5}{2}")
Hope this helps !
Your answer is given below.
Hope this helps !
Answered by
2
kx^2 - 5x + k = 0
the equation has real roots ,it means that
D > 0
==> b^2 - 4ac > 0
==> (-5)^2 - 4×k×k > 0
==> 25 - 4k ^2 > 0
==> 25 > 4k^2
==> 25/4 > k^2
==> 5/2 > k
==> k < 2.5
hope it helps......
==>
the equation has real roots ,it means that
D > 0
==> b^2 - 4ac > 0
==> (-5)^2 - 4×k×k > 0
==> 25 - 4k ^2 > 0
==> 25 > 4k^2
==> 25/4 > k^2
==> 5/2 > k
==> k < 2.5
hope it helps......
==>
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