Quadratic equation variables

Question

Suppose that $3+2\sqrt2$ solves $x^2-6x+a=0$. Find the value of $a$.

• a) $1$
• b) $3$
• c) $5$
• d) $6$

option e) is cut out of the picture and it is $8$.

I want a step by step explanation on how to solve this.

Answer 1

If irrational roots are there for a quadratic then they occur in pairs this can be verified by Descartes rule and by simple quadratic formula x=( -b +/- √(b²-4ac) )/2a if one root is radical then other also is a radical and its conjugate. Now according to your question 3+2√2 is a root so must be 3 - 2√2. Now we know that a equals product of roots in the equation, so simply (3+2√2)(3-2√2) gives your answer which equals 3² - (2√2)² = 1