What causes a solution to a rational equation to be an extraneous solution?

Mathematics

2 answers:

LenKa [72]3 years ago

6 0

its B just did the pre test

nika2105 [10]3 years ago

5 0

Answer: If a solution results in zero when subsitituted into the denominator of the equation, the solution is extraneous.

This is because you obtained the solution from a simplified version of the original equation, but you have to check if the solution obtained is a real solution of the original equation

In the case that when you substitute the solution in the original version the denominator becomes zero, this solution must be rejected, because the division by zero is not defined.

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Change the fraction w-3/w+5 into an equivalent fraction with the denominator w2 + w – 20.

viktelen [127]

ANSWER

$\frac{ {w}^{2} - 7w + 12}{ {w}^{2} + w - 20 }$

EXPLANATION

The given fraction is,

$\frac{w - 3}{w + 5}$

To get an equivalent fraction, we multiply both the numerator and the denominator by the same quantity that will give us

w²+w-20

in the denominator.

This implies that,

$\frac{w - 3}{w + 5} = \frac{(w - 3)(w - 4)}{(w + 5)(w - 4)}$

We multiply out the numerators and denominators using the distributive property to obtain,

$\frac{w - 3}{w + 5} = \frac{ {w}^{2} - 4w - 3w + 12}{ {w}^{2} - 4w + 5w - 20 }$

This simplifies to

$\frac{w - 3}{w + 5} = \frac{ {w}^{2} - 7w + 12}{ {w}^{2} + w - 20 }$

5 0

3 years ago

Find surface area of this regular pyramid

liq [111]

Answer:

189 ft²

Step-by-step explanation:

Here is the formula...

1/2 * 6 * 36 + 81

Hope this helps

7 0

3 years ago