Question

determine the value of k for which quadratic equation(k+5)x2-(2k+3)x+(k-1)=0 has no real roots

Answer 1

Answer:

(k+5)*2-(2k+3)x+(k-1)=o

Step-by-step explanation:

1st solve bracket

2k+10-2kx+3x+k-0

3k+10-2(x

Answer 2

Answer:

$\frac{29}{4} < k$ to have no real roots in the quadratic equation.

Step-by-step explanation:

Given an equation, $(k+5)x^{2} -(2k+3)x+(k-1)=0$ has no real roots.

The condition for o real roots is Discriminant of the quadratic equation is less than Zero ⇒ D< 0

$(k+5)x^{2} -(2k+3)x+(k-1)=0$ From the equation,

a = k+5

b= -(2k+3)

c= k-1

We know that, $D= b^{2} - (4 \times a \times c)$

D <0 ⇒ $b^{2} - (4 \times a \times c) < 0$

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

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

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

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

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

              $4k^{2}- 4k^{2} + 12k -16k+ 20 +9 < 0$

             $-4k +29 < 0$

              $29 < 4k$

             ∴ $\frac{29}{4} < k$

Hence, the value of k should be ⇒ $k > \frac{29}{4}$ , so that the given quadratic equation has no real roots.