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3 years ago

Give the values of a, b, and c needed to write the equation's standard form. (5 + x)(5 - x) = 7

Mathematics

1 answer:

Tpy6a [65]3 years ago

7 0

For this standard for, a = -1, b = 0 and c = 18.

In order to find this, you must first multiply the two parenthesis by each other.

(5 + x)(5 - x)

25 + 5x - 5x - x^2

25 - x^2

Now we need to set it equal to the other side and solve for 0.

25 - x^2 = 7 ----> subtract 7 from both sides.

18 - x^2 = 0 -----> now readjust the order

-x^2 + 18 = 0

So, once we've reached here, the a value is always the coefficient of the x^2 term (-1), b is the value of the coefficient of the x term (0) and c is the constant (18).

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3 years ago

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3 years ago

PLEASE HELP (30 POINTS) Solve the rational equation x/3 = x^2/x + 5 , and check for extraneous solutions.

anygoal [31]

Option C: $x=0$ and $x=\frac{5}{2}$ are the solutions.

Explanation:

The equation is $\frac{x}{3} =\frac{x^{2} }{x+5}$

We shall determine the value of x, by simplifying the equation.

$\begin{aligned} x(x+5) &=3 x^{2} \\ x^{2}+5 x &=3 x^{2} \\ 2 x^{2}-5 x &=0 \\ x(2 x-5) &=0 \end{aligned}$

Thus, $x=0$ and $x=\frac{5}{2}$ are the solutions.

Now, let us check whether the solutions are extraneous solutions.

Let us substitute $x=0$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}&\frac{0}{3}=\frac{0^{2}}{0+5}\\&0=\frac{0}{5}\\&0=0\end{aligned}$

Thus, both sides of the equation are equal.

Hence $x=0$ is a true solution.

Now, Let us substitute $x=\frac{5}{2}$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}\frac{\left(\frac{5}{2}\right)}{3} &=\frac{\left(\frac{5}{2}\right)^{2}}{\left(\frac{5}{2}\right)+5} \\\frac{5}{6} &=\frac{\left(\frac{25}{4}\right)}{\left(\frac{15}{2}\right)} \\\frac{5}{6} &=\frac{5}{6}\end{aligned}$

Thus, both sides of the equation are equal.

Hence, $x=\frac{5}{2}$ is a true solution.

Thus, solutions are not extraneous.

Hence, Option C is the correct answer.

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3 years ago