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grigory [225]
3 years ago
13

Combine like terms. (x+6)(x+6)

Mathematics
2 answers:
ycow [4]3 years ago
5 0

Answer:

(x+6)(x+6)

(x+6)^2

Always remember this identity

(a+b)^2 = a^2 + b^2 + 2ab

So, Now take x=a , 6 =b

Substitute above values

x^2 + 36 +2(x)(6)

=x^2 +36 +12x

=x(x+12) +36

kakasveta [241]3 years ago
4 0

x squared + 12x squared +36

Step-by-step explanation:

x(x+6) + 6(x + 6)

Then further solving

Hope this helps

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What are the roots of this equation? x2-4x+9=0
zysi [14]

Considering the definition of zeros of a function, the zeros of the quadratic function f(x) = x² + 4x +9 do not exist.

<h3>Zeros of a function</h3>

The points where a polynomial function crosses the axis of the independent term (x) represent the so-called zeros of the function.

In summary, the roots or zeros of the quadratic function are those values ​​of x for which the expression is equal to 0. Graphically, the roots correspond to the abscissa of the points where the parabola intersects the x-axis.

In a quadratic function that has the form:

f(x)= ax² + bx + c

the zeros or roots are calculated by:

x1,x2=\frac{-b+-\sqrt{b^{2}-4ac } }{2a}

<h3>This case</h3>

The quadratic function is f(x) = x² + 4x +9

Being:

  • a= 1
  • b= 4
  • c= 9

the zeros or roots are calculated as:

x1=\frac{-4+\sqrt{4^{2}-4x1x9 } }{2x1}

x1=\frac{-4+\sqrt{16-36 } }{2x1}

x1=\frac{-4+\sqrt{-20 } }{2x1}

and

x2=\frac{-4-\sqrt{4^{2}-4x1x9 } }{2x1}

x2=\frac{-4-\sqrt{16-36 } }{2x1}

x2=\frac{-4-\sqrt{-20} }{2x1}

If the content of the root is negative, the root will have no solution within the set of real numbers. Then \sqrt{-20} has no solution.

Finally, the zeros of the quadratic function f(x) = x² + 4x +9 do not exist.

Learn more about the zeros of a quadratic function:

brainly.com/question/842305

brainly.com/question/14477557

#SPJ1

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If he wishes to spilt it evenly there will be less for each person because they all have to give a peice to the other friend.

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