Question

If sum of the roots of an equation is 6 and one of the roots is 3-root 5,then what is the equation?

Answer 1

Step-by-step explanation:

$let \: the \: roots \: are \: \alpha \: \: \: and \: \: \: \: \beta \\ \alpha + \beta = 6 \\ \\ \alpha = 3 \sqrt{5} \\ \\ \alpha + \beta = 6 \\ = > 3 \sqrt{5} + \beta = 6 \\ = > \beta = 6 - 3 \sqrt{5} \\ \\ required \: quadratic \: equation \: is \: \\ \\ {x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0 \\ = > {x}^{2} - (3 \sqrt{5} + 6 - 3 \sqrt{5} )x + 3 \sqrt{5} (6 - 3 \sqrt{5} ) = 0 \\ = > {x}^{2} - 6x + 18 \sqrt{5} - 9 \sqrt{25} = 0 \\ = > {x}^{2} - 6x + 18 \sqrt{5} - 9 \times 5 = 0 \\ = > {x}^{2} - 6x - 45 + 18 \sqrt{5} = 0$

Answer 2

Answer:

x2-6x+9=0

Step-by-step explanation:

r1+r2 = 6

r2 = 3-root5

r1 = 3+root 5

alpha + beta = -b/a

alpha*beta = c/a

x2-6x+9=0