Hold on

x² + ax + b = 0, a and b are integers.

If 3 + 5√2 is a root of the equation, then x = 3 + 5√2 must satisfy the equation.

Sub in x = 3 + 5√2,

(3 + 5√2)² + a(3 + 5√2) + b = 0

9 + 30√2 + 50 + 3a + 5a√2 + b = 0

59 + b + 3a + (30 + 5a)√2 = 0

If you look at the right side, there's no term in √2 and the integer is 0

So,

59 + b + 3a + (30 + 5a)√2 = 0 + 0√2

Comparing both sides,

59 + b + 3a = 0 ①
30 + 5a = 0 ②

From ②,
5a = -30
a = -6

Sub a = -6 into ①,

59 + b + 3(-6) = 0
b + 41 = 0
b = -41


The original equation is :

x² - 6x - 41 = 0

Edited.

There is actually a shortcut to this question.

If one root is 3 + 5√2, then the other root must be its conjugate, 3 - 5√2.

Then 3 + 5√2 is something you will expect when you use the quadratic formula or complete the square for those equations that do not have rational roots.

So,

x = 3 + 5√2 or x = 3 - 5√2

x - (3 + 5√2) = 0 or x - (3 - 5√2) = 0

Multiply both together and you will still get 0.


(x - (3 + 5√2))(x - (3 - 5√2)) = 0

x² - (3 - 5√2)x - (3 + 5√2)x + (3² - (5√2)²) = 0

x² - 3x + 5√2 x - 3x - 5√2 x + (9 - 50) = 0

x² - 6x - 41 = 0

a = -6, b = -41

Alternatively, even shorter way :

Sum of roots = 3 + 5√2 + 3 - 5√2 = 6

Product of roots = (3 + 5√2)(3 - 5√2)

= 3² - (5√2)²

= 9 - 50

= -41

Equation is : x² - (sum)x + product = 0

x² - 6x - 41 = 0

As we can see,

a = -(sum) = -6
b = product = -41

Why? Because recall that sum of roots = -b/a and product of roots = c/a for an equation in the form ax² + bx + c = 0

So for this equation x² + ax + b,

Sum = -a/1 = -a
Product = b/1 = b

Why is the 2nd method b=41 and not -41

Edited already before your comment

Oh sorry,can help me with one last question

Idt can use method 3 as the question ask us not to use a calculator so how will we be able to get 6 without a calculator

Why do you need a calculator for it?

3 + 5√2 + (3 - 5√2)

= 3 + 5√2 + 3 - 5√2

= 3 + 3 + 5√2 - 5√2

= 6

The two 5√2s have already cancelled out.

Oh ya sorry my mistake didn't calculate it out

I just thought of another method to do this question :


x² + ax + b = 0

Using the quadratic formula,

x = (-a ± √(a² - 4(1)(b) ) / 2(1)

x = -½a ± ½√(a² - 4b)

Since 3 + 5√2 is a root of the equation,

Then 3 + 5√2 = -½a + ½√(a² - 4b)

3 + √(2 x 5²) = -½a + √(¼(a² - 4b))

3 + √50 = -½a + √(¼a² - b)

Comparing both sides,

3 = -½a → a = -6

And

50 = ¼a² - b

b = ¼(a²) - 50

= ¼(-6)² - 50
= ¼(36) - 50
= 9 - 50
= -41

Will help you with the newer qn now

Oh you using the half coefficient of X method

Basically, quadratic formula and then comparing coefficients.

a and b are given as integers so for sure that 3 = -a/2

The square root(irrational number term) on the RHS has to equal the 5√2 on the LHS. So just make both into a single square root term and compare the inside.

"If one root is 3 + 5√2, then the other root must be its conjugate, 3 - 5√2"

I'm pretty certain a large majority of O Level students do not know this property or at least do not pay proper attention to the format of the quadratic equation when they solve quadratic equation which yields irrational figures.

That's the whole point of showing this method.

To get them to realise it.

Who knows, it may be part of some higher order exam question that tests conceptual understanding.