Question

Find the zeroes of polynomial p(x)= x^2 +2√2x -6

Answer 1

$\textbf{Hey!!}$

Given polynomial

$p(x) = x {}^{2} + 2 \sqrt{2} x - 6$

On comparing with ax^2 + bx + c we get,

a = 1 , b = 2√2 , c = - 6

Sum of zeroes

$\alpha + \beta = \frac{ - b}{a} \\ \\ = > \alpha + \beta = \frac{ - 2 \sqrt{2} }{1} \\ \\ = > \alpha + \beta = - 2 \sqrt{2}$

Product of zeroes

$\alpha \beta = \frac{c}{a} \\ \\ = > \alpha \beta = \frac{ - 6}{1} \\ \\ = > \alpha \beta = - 6$

Required polynomial

$= k(x {}^{2} - ( \alpha + \beta )x + \alpha \beta ) \\ \\ = k(x { }^{2} - ( - 2 \sqrt{2} )x + ( - 6) \\ \\ = k(x {}^{2} + 2 \sqrt{2} x - 6 \\ \\ where \: \: k \: \: is \: \: constant \\ \\ = x {}^{2} + 2 \sqrt{2} x - 6$

Therefore the required polynomial is

$x {}^{2} + 2 \sqrt{2}x - 6.$

$\textbf{Hope it helps!}$

Answer 2

hey mate here's ur answer,
x² + 2√2x - 6
now, comparing the given polynomial with
ax² + bx + c, we get
a = 1,b = 2√2, c = -6
D = b² - 4ac
= (2√2)² - 4*1*(-6)
= 8 + 24 = 32
x = (-b +- √D)/(2a)
x = (-2√2 +-√32)/2
x = (-2√2 + 4√2)/2, (-2√2 - 4√2)/2
x = 2(-√2 + 2√2)/2, 2(-√2 - 2√2)/2
x = -√2 + 2√2, -√2 - 2√2
x = √2(-1 + 2), -√2(1 + 2)
x = √2, -3√2
these are the roots of the given polynomial.