Question

find k and the other root if x²-6x +K=0
has 3 + √2 as one root.​

Answer 1

Answer:

k=7

Step-by-step explanation:

x²-6x +K=0

(3+√2)²-6(3+√2)+k=0

9+6√2+2-18-6√2+k=0

9+2-18+6√2-6√2+k=0

-7+k=0

k=7

Answer 2

Answer:

$k = 7 \: and \: other \: root = (3-√2)$

Step-by-step explanation:

$put \: x \: = \: (3+ √2) \: in \: the \: equation.</p><p></p><p> \\ (3+√2)²-6(3+√2)+ k = 0 \\ </p><p>or, 9 + 2 + 6√2 -18 -6√2 + k= 0 \\ </p><p>or, -7 + k = 0 \\ </p><p>or, k = 7. \\ </p><p></p><p>put \: k= 7 \\ </p><p></p><p> x²-6x + 7= 0 \\ </p><p></p><p>D = b²-4ac = (-6)² - 4.1.7 = 36 - 28= </p><p></p><h3>D =8</h3><p>$

$x = \frac{- b ± \sqrt{ {b}^{2} - 4ac } }{2a}$

$x = \frac{6± \sqrt{8} }{2} \\ = \frac{2(3± \sqrt{2}) }{2} \\ = 3± \sqrt{2}$

$Hence, \: the \: roots \: are \: (3+√2) \: and \:</p><p> (3-√2).$