I only need (a) ^ ^

Two ways to do this :

Method 1

5 - √3 is a root, so substitute it back into the equation. (i.e x = 5 - √3)


-(5 - √3)² + a(5 - √3) + b = 0

-(5² - 2(5)(√3) + (√3)²) + 5a - a√3 + b = 0

-(25 - 10√3 + 3) + 5a - a√3 + b = 0

10√3 - 28 + 5a - a√3 + b = 0

5a + b - a√3 = 28 - 10√3


Recall that a and b are integers.

So, comparing the coefficients of the integers on both sides,

5a + b = 28 ①

Comparing the coefficients of the surds on both sides,

a = 10

Substitute a = 10 into ①,

5(10) + b = 28
50 + b = 28
b = 28 - 50
b = -22

Method 2

The Conjugate Root Theorem :

If P is a polynomial in one variable with real coefficients, and a + bi is a root of P with a and b real numbers, then its complex conjugate a − bi is also a root of P.

This means that complex roots occur in pairs.


Likewise, this extends to irrational roots. (Irrational Root Theorem)


If a + √b is a root of the polynomial, then a - √b is also a root.


So if 5 - √3 is a root, then 5 + √3 is also a root.


So, work backwards from how you usually solve quadratic equations.


x = 5 - √3 or x = 5 + √3

x - (5 - √3) = 0 or x - (5 + √3) = 0

(x - (5 - √3))(x - (5 + √3)) = 0

x² - (5 + √3)x - (5 - √3)x + (5 - √3)(5 + √3) = 0

x² - (5 + √3 + 5 - √3)x + 5² - (√3)² = 0

x² - 10x + 22 = 0

-x² + 10x - 22 = 0

a = 10, b = -22

Alternatively,

Product of roots

= (5 - √3)(5 + √3)
= 5² - (√3)²
= 25 - 3
= 22

Sum of roots

= 5 - √3 + 5 + √3
= 10


Recall that the for a quadratic equation in the form ax² + bx + c = 0,

Sum of roots = -b/a
Product of roots = c/a


So, for -x² + ax + b = 0,

Sum of roots = -a/(-1) = a
So a = 10


Product of roots = b/(-1) = -b
So 22 = -b
b = -22



Or


Quadratic equation :

x² - (sum of roots)x + product of roots = 0

x² - 10x + 22 = 0
-x² + 10x - 22 = 0

a = 10 , b = -22