Question

find the value of b, if x=-2 is a root of equation x²+2bx+6=0​

Answer 1

Answer:

putting the value of X in equation x^2 + 2bx + 6

Step-by-step explanation:

x^2 + 2bx +6

(-2)^2 + 2b × -2 + 6

4 + (-4b) + 6 = 0

10 - 4b = 0

-4b = -10

b = -10/-4

b = 5/2 is correct answer

Answer 2

Question:

Find the value of b, if x = -2 is a root of equation x² + 2bx + 6 = 0 .

To find:

★ The value of b

Answer:

$The \: value \: of \: \underline{ \:\underline{ \: \sf \pink{\bold{b \: is \: \frac{5}{2}} } \: }\: }$

Given:

$\star Root \: of \: equation \: \underline{\bold{ x \: is \: - 2} }$

$\star An \: equation \: is \: given,$

${x}^{2} + 2bx + 6 = 0 \: --->(1)$

Step-by-step explanation:

Now substituting the value of x in eq(1)

${x}^{2} + 2bx + 6 = 0$

$\implies {( - 2)}^{2} + 2(b)( - 2) + 6 = 0$

$\implies 4 - 4b + 6 = 0$

$\implies - 4b + 10 = 0$

$\implies 4b = 10$

$\implies 2b = 5$

$\implies \boxed{b = \frac{5}{2} }$

$\therefore The \: value \: of \: \underline{ \:\underline{ \: \bold{b \: is \: \frac{5}{2} } \: }\: }$

Verification:

Now substituting the value of x and b in eq ( 1 )

$\implies {( - 2)}^{2} + \not{ 2}( \frac{5}{ \not{2}})( - 2) + 6 = 0$

$\implies 4 - 10 + 6 = 0$

$\implies 10 - 10 = 0$

$\implies \boxed{ 0 = 0}$

$\underline{L.H.S.=R.H.S.}$

$Hence \: verified.$