Question

If x^2 + bx + b + 3 = 0 has roots of the form of (-b+√5)/2, where b > 0, then b = m+√n for positive integers m,n. Find m + n.

Answer 1

Answer:

23

Step-by-step explanation:

Using quadratic formula :

In eq. x^2 + bx + ( b + 3 ) = 0

Discriminant = b^2 - 4( b + 3 ) = b^2 - 4b - 12.

x = [ - b ± √( discriminant ) ] / 2

x = [ - b ± √( b^2 - 4b - 12 ) ] / 2

Given equation has roots ( - b ± √5 ) / 2

= > ( - b ± √5 ) / 2 = [ - b ± √( b^2 - 4b + 3 ) ] / 2a

= > √5 = √( b^2 - 4b -12 )

= > 5 = b^2 - 4b - 12

= > b^2 - 4b - 17 = 0

Using quadratic formula :

= > b = [ 4 ± √( 16 + 68 ) ] / 2

= ( 4 ± √84 ) / 2

= ( 4 ± 2√21 ) / 2

= 2 ± √21

As given, b = m + √n, and from the above result :

= > m + √n = 2 + √21

Comparing values : m = 2 & n = 21

Hence, m + n = 2 + 21 = 23

Answer 2

Answer:

23

Step-by-step explanation:

Using the quadratic formula, we see that the solutions of the quadratic equation $x^2 + bx + (b+3) = 0$ are given by $\frac{-b \pm \sqrt{b^2 - 4(b+3)}}{2}$. So we may set $\frac{-b + \sqrt{b^2 - 4(b+3)}}{2}$ equal to $\frac{-b+\sqrt{5}}{2}$ which implies $b^2 - 4b - 12 = 5 \Longrightarrow b^2 - 4b - 17 = 0$. (Note that setting $\frac{-b + \sqrt{b^2 - 4(b+3)}}{2}$ equal to $\frac{-b-\sqrt{5}}{2}$ gives no solution). We must use the quadratic formula again. We get$$b = \frac{4 \pm \sqrt{4^2 - 4(-17)}}{2} = \frac{4 \pm \sqrt{84}}{2} = 2 \pm \sqrt{21}.$$Take the positive root and sum: $m+n = 2+21 = \boxed{23}$.

You can put what I said into latex and see the explanation