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svlad2 [7]
1 year ago
5

Solve the radical equation. Check all purpose solutions. Sqrt(4-3x) = x

Mathematics
1 answer:
alisha [4.7K]1 year ago
5 0

Step-by-step explanation:

√(4-3x) = x

1. square both sides:

[√(4-3x)]²= (x) ²

4-3x = x²

2. shift all the terms to the right side

(minimise dealing with negative coefficient of x² terms to prevent careless mistakes)

0= x²+3x-4

no matter how you change the places of the terms, it will still be:

x²+3x-4=0

3. solve the quadratic equation, either by cross multiplication(and factorising) or the quadratic formula(if you're taught)

for me I will be using the cross multiplication table that my calculator can help me with(gives me the solution but better to show the factorised form to prevent missing out method marks)

as such:

x²+3x-4=0

(x+4)(x-1)=0

hence:

(x+4)=0 or (x-1)=0

finally:

x= -4 or x= 1

however we're not done yet as some solutions may not match with the question/ original equation provided, this we still need to check if there are any values that does not match

substitute x= -4 into the original equation:

√4-3(-4) = √4+12 = √16 = 4 > -1 , hence x= -1 is not the answer

you can substitute the values into a scientific calculator(that has the functions required) to quickly determine

then to double confirm, do the same for the other value of X that we have found:

√4-3(1) = √4-3 = √1 = 1

since the other value of x matches with the equation provided in the question, then we can now confirm that the value of x=1 is the correct value

reason for such situation to happen:

remember in step 1 when we squared both sides of the equation, that is the reason for having to check for the correct X value. let me explain further

when we square numbers, we are potentially introducing another possible answer other than the correct answer intended for us to find as well as to match with the question.

for example:

1 = 1² = (-1)²

likewise, I can even take any other number to compare, such as 100:

100=10²=(-10)²

and this scenario only happens for the even powers of x, such as x², x⁴, x⁶ so on and so forth.

hence whenever such questions come out, do remember to double check with the original question/equation

lastly, the checking portion will only apply for even powers of x in questions and/or when you happen to square them

hope this helps!

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Step-by-step explanation:

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Answer:

  • Domain: All the real values except x = 2 and x = 4: R - {2, 4}
  • Holes: x = 2
  • VA, vertical asymptores: x = 4
  • HA: horizontal asymptotes: there are not horizontal asymptotes
  • OA: oblique asymptotes: x + 6 [note that OH does not stand for any known feature, and so it is understood that it was intended to write OA]
  • Roots: x = 2
  • Y-intercept: -1

Step-by-step explanation:

1. <u>Given</u>:

f(x)=\frac{x^3-8}{x^2-6x+8}

  • Note that the number 8 in the numerator is not part of the power.
  • Type of function: rational function

2. <u>Domain</u>: is the set of x-values for which the function is defined.

The given function is defined for all x except those for which the denominator equals 0.

  • Denominator:  x² -6x + 8 = 0
  • Solve for x:

        Factor. (x - 4 )(x - 2) = 0

        Zero product property: (x - 4) = 0 or (x - 2) = 0

        x - 4 = 0 ⇒ x = 4

        x - 2 = 0 ⇒ x = 2

  • Domain:

        All the real values except x = 2 and x = 4: x ∈ R / x ≠ 2 and x ≠ 4.

3. <u>Holes</u>:

The holes on the graph of a rational function are at those x-values for which both the numerator and denominator are zero.

  • Find the values for which the numerator is zero:

        Numerator: x³ - 8 = 0

        Factor using difference of cubes property:

                   a³ - b³ = (a - b)(a² + ab + b²)

                   x³ - 8 = (x - 2)(x² + 2x + 4) = 0

        Zero product property:  (x - 2)(x² + 2x + 4) = 0

                    x - 2 = 0 ⇒ x = 2                    

                    x² + 2x + 4 = 0 (this has not real solution)

  • The values for which the denominator is zero were determined above: x = 2 and x = 4.

  • Conclusion: for x = 2 both numerator and denominator equal 0, so this is a hole.

4. <u>VA: Vertical asymptotes</u>.

The vertical asymptotes on the graph of a rational function are the vertical lines for which only the denominator (and not the numerator) equals zero.

  • In the previous part it was determined that happens when x = 4.

5. <u>HA: Horizontal asymptotes</u>.

In rational functions, if the numerator is a higher degree polynomial than the denominator, there is no horizontal asymptote.

6. <u>OA: oblique asymptotes</u>

  • Find the quotient and the remainder.

                       x + 6

                  _______________

x² - 6x + 8 )   x³ + 0x² + 0x - 8

                  - x³ + 6x² - 8x

                   ___________

                          6 x² -   8x -  8

                        - 6x² + 36x - 48

                        _____________

                                    28x  - 56

Result: (x + 6) + (28x - 56) / (x² - 6x + 8)

  • Find limit x → ∞

\lim_{x \to \infty}(x + 6) + \frac{28x-56}{x^2-6x+8}=x+6

<u>7. Roots</u>:

Roots are the values for which f(x) = 0.

That happens when the numerator equals 0, and the denominator is not 0.

As determined earlier: x³ - 8 = 0 ⇒ x = 2.

8. <u>Y-Intercept</u>

The y-intercepts of any function are the y-values when x = 0

  • Substitute x = 0 into the function:

         f(x)=\frac{x^3-8}{x^2-6x+8}=\frac{0^3-8}{0^2-6(0)+8}}=\frac{-8}{8} =-1

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Step-by-step explanation:

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