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nekit [7.7K]
4 years ago
12

What are the zeros of the polynomial function? F(x)= x^2 +12x

Mathematics
1 answer:
Pavel [41]4 years ago
8 0

Answer:

x = 0 and x = -12 are the two zeroes

Step-by-step explanation:

"Zeroes" of a function are also called the solutions, roots or x intercepts.

The zeroes are the numbers that give the function a values of 0, so find what value for 'x' yields F(x) = 0 as a result.

With quadratic equations (equations with x² as the largest exponent), you factor the equation.  

Step 1: Set the equation equal to zero

  x² + 12x = 0

 Depending on the equation, there are several methods you can use to factor, for this one we can factor out an 'x' since both terms have an 'x' in them.  We get...

  x(x + 12) = 0      

So now, using the "zero product property" which means that if two things are multiplied together, and the result is zero, then either the first term is zero, the second term is zero, or they are both zero.  So we set each term equal to zero and solve.

So

either    x = 0     or        x + 12 = 0

Solve each for x to get your two anwers.

x= 0 is already solved, so 0 is our first zero

Now solve

x + 12 = 0  

  x = -12             (subtract 12 from both sides to isolate x)

 

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Line e has a slope of 7/4 line f is parallel to line e what is the slope for line f
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4 years ago
) Rearrange v = u + at to make t the subject of the formula.
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Question:-
8090 [49]

Step-by-step explanation:

\large \rm \red{\widetilde{Taking\ RHS:-}}

\rm :\longmapsto \dfrac{1}{3 + \sqrt{7}} \ + \ \dfrac{1}{\sqrt{7} + \sqrt{5}} \ + \ \dfrac{1}{\sqrt{5} + \sqrt{3}} \ + \ \dfrac{1}{\sqrt{3} + 1}

<u>Rationalizing, we get</u>

\rm :\longmapsto \dfrac{1}{3+\sqrt7} \times \dfrac{3-\sqrt7}{3-\sqrt7} + \dfrac{1}{\sqrt7 + \sqrt5} \times \dfrac{\sqrt7 - \sqrt5}{\sqrt7 - \sqrt5}

\rm + \dfrac{1}{\sqrt5 + \sqrt3} \times \dfrac{\sqrt5 - \sqrt3}{\sqrt5-\sqrt3} + \dfrac{1}{\sqrt3 + 1} \times \dfrac{\sqrt3 - 1}{\sqrt3 -1}

\rm :\longmapsto \dfrac{3 - \sqrt7}{(3)^2 - (\sqrt7)^2}

+ \dfrac{\sqrt7 - \sqrt5}{(\sqrt7)^2 - (\sqrt5)^2}

+\dfrac{\sqrt5 - \sqrt3}{(\sqrt5)^2 - (\sqrt3)^2} +

\dfrac{\sqrt3-1}{(\sqrt3)^2 - 1^2}

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3 years ago
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