Question

if one of the root of x^2+bx+c=0(where b,c are rational numbers) is 2-√3, then b+c is equal to​

Answer 1

Step-by-step explanation:

if one of the root is 2-√3

then other root must be 2+√3

so sum of the zeros = -b/1

(2-√3)+(2+√3)=-b

b=-4

and product of the zeros = c/1

(2-√3)(2+√3)=c

(4-3)=c

c=1

so b+c= -4+1=-3

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:1 \: root \: of \: {x}^{2} + bx + c = 0 \: is \: 2 - \sqrt{3}$

We know, that irrational roots occur in conjugate pairs.

So, Let assume that roots of the equation be

$\rm :\longmapsto\: \alpha = 2 - \sqrt{3}$

and

$\rm :\longmapsto\: \beta = 2 + \sqrt{3}$

We know, that,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\bf\implies \: \alpha + \beta = - \: \dfrac{b}{1}$

$\rm :\longmapsto\:2 - \sqrt{3} + 2 + \sqrt{3} = - b$

$\rm :\longmapsto\:4= - b$

$\rm \implies\:\boxed{ \tt{ \: b \: = \: - \: 4 \: \: }}$

Now, also we know that

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\: \alpha \beta = \dfrac{c}{1}$

$\rm :\longmapsto\:c = (2 - \sqrt{3})(2 + \sqrt{3})$

$\rm :\longmapsto\:c = {2}^{2} - {( \sqrt{3} )}^{2}$

$\rm :\longmapsto\:c = 4 - 3$

$\rm \implies\:\boxed{ \tt{ \: c \: = \: 1\: \: }}$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: b + c = \: - \: 4 + 1 \: = \: - \: 3 \: }}}$

Additional Information :-

1. For cubic equation

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: roots \: of \: a {x}^{3} + b {x}^{2} + cx + d = 0, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$