Question

9) Show that the equation x2 + ax -1 = 0 has real and distinct roots for all
real values of a​

Answer 1

Answer:

Step-by-step explanation:

We know that for having real and distinct roots $\sqrt{bx^{2} - 4ac}$ > 0

So, in our given equation b= a, a= 1 and c= -1

putting them in the above equation:-

$\sqrt{a^{2}-4(1)(-1) }$ >0

$\sqrt{a^{2} +4}$ >0

Hence for all real values of a the equation will have real and distinct roots

Answer 2

Step-by-step explanation:

D=b²−4ac

$D = a {}^{2} - 4(1)(1) = a² - 4$

for all the real values of a>2and a<−2

$D >0$

hence x² + ax + 1 = 0 has real distinct roots for all the real values of a.