Respuesta :
So, we need to solve the quadratic equation from quadratic formula as per the question, but before starting, let's first recall that, for any quadratic equation of the form ax² + bc + c = 0, the discriminant is given by D = b² - 4ac, and the value of x is given by :
- [tex]{\boxed{\bf{x=\dfrac{-b\pm \sqrt{D}}{2a}}}}[/tex]
Also, their are no real solutions of x i.e their are only imaginary values of x iff D < 0, equal and real roots iff D = 0 and real and distinct roots iff D > 0. So, let's convert the given equation to the standard form ax² + bx + c = 0 and then we will find the discriminant and then we will be solving for k ;
[tex]{:\implies \quad \sf 2k(k+5)=-25}[/tex]
[tex]{:\implies \quad \sf 2k^{2}+10k=-25}[/tex]
[tex]{:\implies \quad \sf 2k^{2}+10k+25=0}[/tex]
Now, on comparing this with the standard equation, we will get a = 2, b = 10 and c = 25, and then D = (10)² - 4 × 2 × 25 = 100 - 200 = - 100, now as D < 0, so their will be imaginary values of k, so now solving for k will yield by using the Quadratic Formula ;
[tex]{:\implies \quad \sf k=\dfrac{-10\pm \sqrt{-100}}{2\times 2}}[/tex]
[tex]{:\implies \quad \sf k=\dfrac{-10\pm 10\iota}{2\times 2}}[/tex]
[tex]{:\implies \quad \sf k=\dfrac{2(-5\pm 5\iota)}{2\times 2}=\dfrac{-5\pm 5\iota}{2}}[/tex]
[tex]{:\implies \quad \boxed{\bf{k=\dfrac{-5+5\iota}{2}\:\:,\:\: \dfrac{-5-5\iota}{2}}}}[/tex]
This is the required answer
