Question

The quadratic equation having roots (2 + √3 ) and (2 ─ √3 ) will be : *

1 point

(a) x^2+ 4x+1 = 0

(b) x^2- 4x+1 = 0

(c) x^2+ 4x-1 = 0

(d) x^2- 4x-1 = 0

​

Answer 1

Answer:

answer b

Step-by-step explanation:

sum of the roots = 2+√3+2-√3=4

product of the roots = (2+√3)(2-√3)=4-3=1

if we know sum and product then equation of quadratic equation =

x^2-x(sum of the roots) +product of the roots =0

so

our equation is

x^2-4x+1=0

Answer 2

GIVEN :-

• Roots of quadratic equation are (2 + √3) and (2 - √3).

TO FIND :-

• The quadratic equation.

SOLUTION :-

As we know that the quadratic equation is given by,

$\\ : \implies \displaystyle \sf \: Polynomial = x ^{2} - ( \alpha + \beta )x + ( \alpha \beta )$

• α = (2 + √3)
• β = (2 - √3)

$\\ \\ : \implies \displaystyle \sf \: Polynomial =x ^{2} - \bigg[(2 + \sqrt{3} + 2 - \sqrt{3} )x + (2 + \sqrt{3})(2 - \sqrt{3} )\bigg]$

$: \implies \displaystyle \sf \: Polynomial = x ^{2} - ( 2 + 2)x + (2) ^{2} - ( \sqrt{3} ) ^{2}$

$: \implies \displaystyle \sf \: Polynomial = x ^{2} - 4x + 4 - 3$

$: \implies \underline{ \boxed{\displaystyle \sf \bf{\: Polynomial = x ^{2} - 4x + 1=0}}}$