What is the proper way to check an algebraic solution?

2 answers:

6 0

Substitution is a great way to find out if the statement or equation is true and if it is both sides will be equal to the same number. Now in the case of any solution I'd just check with random numbers about 2 times. If it's no solution do the same thing. And with no solution or any solution be sure to double check your work.

vlabodo [156]4 years ago

4 0

The correct way is to plug in the answer back into the equation and see if it results in the correct solution.

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WARRIOR [948]

Answer:

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Step-by-step explanation:

6 0

3 years ago

Somebody please help so I can pass, please

ASHA 777 [7]

First, we are going to find the vertex of our quadratic. Remember that to find the vertex $(h,k)$ of a quadratic equation of the form $y=a x^{2} +bx+c$ , we use the vertex formula $h= \frac{-b}{2a}$ , and then, we evaluate our equation at $h$ to find $k$ .

We now from our quadratic that $a=2$ and $b=-32$ , so lets use our formula:
$h= \frac{-b}{2a}$
$h= \frac{-(-32)}{2(2)}$
$h= \frac{32}{4}$
$h=8$
Now we can evaluate our quadratic at 8 to find $k$ :
$k=2(8)^2-32(8)+56$
$k=2(64)-256+56$
$k=128-200$
$k=-72$
So the vertex of our function is (8,-72)

Next, we are going to use the vertex to rewrite our quadratic equation:
$y=a(x-h)^2+k$
$y=2(x-8)^2+(-72)$
$y=2(x-8)^2-72$
The x-coordinate of the minimum will be the x-coordinate of the vertex; in other words: 8.

We can conclude that:
The rewritten equation is $y=2(x-8)^2-72$
The x-coordinate of the minimum is 8