How do you find the discriminant, describe the number and type of root, and find the exact solution using the quadratic formula given #2x-5=-x^2#?

Question

Christian Antunez · Accepted Answer

Roots are #-1+sqrt6# and #-1-sqrt6#

For an equation #ax^2+bx+c=0#, discriminant is #b^2-4ac#

and quadratic formula gives roots as #x=(-b+-sqrt(b^2-4ac))/(2a)#

Note that if #a,b#and #c# are rational number, as we have in the given example

if #b^2-4ac# is a perfect square (i.e. positive as well), roots are rational

if #b^2-4ac# is positive but not a perfect square, roots are irrational and conjugate i.e. of the type #p+-sqrtq#

if #b^2-4ac# is negative, roots are complex conjugate i.e. of the type #p+-iq#

Here we have #2x-5=-x^2# i.e. #x^2+2x-5=0#

and as #a=1#, #b=2# and #c=-5#, the discriminant is

#b^2-4ac=2^2-4xx1xx(-5)=4+20=24#

and hence roots are irrational and conjugate.

These are #(-2+-sqrt24)/2=-2/2+-(2sqrt6)/2=-1+-sqrt6#

Lincoln Anderson · Answer

undefined To find the discriminant, number and type of roots, and the exact solution using the quadratic formula for the equation $2x - 5 = -x^2$:

1. Begin by rewriting the equation in standard form: $x^2 + 2x - 5 = 0$.
2. Identify the coefficients: $a = 1$, $b = 2$, and $c = -5$.
3. Calculate the discriminant using the formula $\Delta = b^2 - 4ac$.
4. Substitute the values: $\Delta = (2)^2 - 4(1)(-5)$.
5. Simplify to find $\Delta$.
6. Determine the nature of the roots:
   - If $\Delta > 0$, the equation has two distinct real roots.
   - If $\Delta = 0$, the equation has one real root (a repeated root).
   - If $\Delta < 0$, the equation has two complex roots.
7. Use the quadratic formula to find the exact solutions:
   - $x = \frac{{-b \pm \sqrt{\Delta}}}{{2a}}$.
   - Substitute the values of $a$, $b$, and $\Delta$.
   - Simplify to find the exact solutions for $x$.