Question

Use expansion formula and solve(6+2x)²​

Answer 1

Step-by-step explanation:

(6+2x)²

(6)²+(2x)²+2×6×2x. ; [a+b]²= a²+b²+2ab

4x²+24x+36

4x²+12x+12x+36=0

4x(x+3)+12(x+3)=0

(4x+12)(x+3)=0

x= –12/4. or x= –3

Answer 2

Hey Buddy,

Thanks For Asking the Question, Your Answer is below with Steps

Explanation:-

(6+2x)² can be written as (2x+6)²

Using Binomial expansion Formula :-

$(a+b)^{n} =$ ∑  n  and k=0   $\left \{ {{n} \atop k}} \right }$  }  $a^{n-k} b^{k}$

where $\left \{ {{n} \atop k}} \right }$  }  = $\frac{n!}{(n-k)!k!}$ and n! = 1 ⋅ 2 ⋅ . . . . . . . . ⋅ n

we have that a = 2x , b = 6 and n = 2

∴ ( 2x + 6 )² = ∑ 2 and k=0   $\left \{ {{2} \atop k }$  }  $(2x)^{2-k} 6^{k}$

Now, calculate the product for every value of k from 0 to 2.

k = 0 :   $\left \{ {{2} \atop 0 }$   $(2x)^{2-0}$ . ( dot product ) $6^{0}$ =  $\frac{2!}{(2-0)!0! }(2x)^{2-0} .\ 6^{0} =4x^{2}$

k = 1 :   $\left \{ {{2} \atop 1 }$    $(2x)^{2-1}.\ 6^{1}$ = $\frac{2!}{(2-1)!1!} (2x)^{2-1} .\ 6^{1} = 2\ . \ 2x\ . \ 6 = 24x$

k = 2 :  $\left \{ {{2} \atop 2 }$    $(2x)^{2-2}.\ 6^{2}$  = $\frac{2!}{(2-2)!2!} \ (2x)^{2-2} \ . \ 6^{2} = 1\ . \ 36 = 36$

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                          ANSWER

                 $(2x+6)^{2}\ = \ 4x^{2} \ + \ 24x \ + \ 36$

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Finding Roots of $4x^{2} +24x+36$

Taking 4 common in the expression. we get,

⇒ $4 (x^{2} +6x+ 9)$

⇒ $4(x^{2} +3x+3x+9)$

⇒ $4 ( x + 3 )\ (x+3)$

⇒ $4 ( x + 3 )^{2}$

                                    $\hug{\ornage{\boxed{\boxed{\pink{\mathscr{ x = -3}}} }} }}}$

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Hope It Helps You Dear ! :D    ^_^

By Abhiram5566

                                 MARK ME AS BRAINLIEST ANSWER

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