Question

consider the equation
(k+5) x2-(2k+3)x+k-1=0
for what value f
of k does the equation has no real roots​

Answer 1

Answer:

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Answer 2

k > $\frac{29}{4}$

Given equation is (k+5)$x^{2}$-(2k+3)+k-1

On compairing the equation with $ax^{2} +bx+c =0$

a= k+5 b=-2k-3 c= k-1

The equation has no real roots when Discriminant < 0

i.e. $b^{2}-4ac < 0$

$(-2k-3)^{2} - 4(k+5)(k-1) < 0$

$(-2k)^{2} + (-3)^{2} +2(-2k)(-3) -4 (k^{2} +5k-k-5) < 0$$4k^{2} + 9 + 12k -4( k^{2} +4k-5) < 0$

$4k^{2} +9 +12k-4k^{2} - 16k+20 < 0$ (solving the like terms )

$-4k+29 < 0$ (adding 4k to both sides )

29< 4k ( dividing 4 from both the sides )

$\frac{29}{4 } < k$