How do you solve #x^2-5x-8=0#?

Answer 1

The roots are #6.275, -1.275#

This is a quadratic equation of form #ax^2+bx+c# where #a=1;b=-5;c=-8 ; b^2-4ac=25+32=57(+)#ive. So roots are real. #x= -b/(2a) +- sqrt (b^2-4ac)/(2a)= 2.5 +- sqrt57/2 = 6.275, -1.275# [Ans]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Savannah Anderson

You can solve the equation (x^2 - 5x - 8 = 0) using the quadratic formula:

[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}]

Where (a = 1), (b = -5), and (c = -8). Plugging in these values:

[x = \frac{{-(-5) \pm \sqrt{{(-5)^2 - 4(1)(-8)}}}}{{2(1)}}]

[x = \frac{{5 \pm \sqrt{{25 + 32}}}}{{2}}]

[x = \frac{{5 \pm \sqrt{{57}}}}{{2}}]

So the solutions are (x = \frac{{5 + \sqrt{57}}}{{2}}) and (x = \frac{{5 - \sqrt{57}}}{{2}}).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.