Nieleaidn
Answered

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1.f(x)=x²-4x+4
2.f(x)=x²-7
3.f(x)=x²+6+9
4.f(x)=x²+10x
 these are quadratic functions. guys please help me, im having a hard time in solving these... or  just explain me how to put this in standard form. it would mean alot to me..

Sagot :

Rider21idn
The standard form of a quadratic Equation takes this form:
 
                          Ax² + Bx + C = 0
       where A,B, and C are numerical coefficients. A should not be equal to zero..

Let f(x) = 0.

So we will have the following in standard form:

1.) x² - 4x + 4 = 0
2.) x² - 7 = 0
3.) x² + 15 = 0
4.) x² + 10x = 0

As you noticed, all the given are in standard form already..except that we just simplified #3.

If you were asking for the solutions/roots of each equation:

1.) x² - 4x + 4 = 0
    (x-2)(x-2) = 0
     x₁ and x₂ are both equal to 2.......so 2 is the one and only root to the equation.

2.) x² - 7 = 0
    x = +/- √7
    x₁ = +√7            x₂ = -√7            ANSWERS

3.) x² + 15 = 0
    x² = -15
   x² = √-15
since there is no square root of a negative number, the equation has NO REAL ROOTS.

4.) x² + 10x = 0
     x(x + 10) = 0

x₁ = 0  ANSWER         and  x+10 = 0
                           x = -10  ANSWER

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