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Avery solves the equation below by first squaring both sides of the equation.
√z^2+8 = 1 -2z
What extraneous solution does Avery obtain?
z= ?


Avery Solves The Equation Below By First Squaring Both Sides Of The Equation Z28 1 2z What Extraneous Solution Does Avery Obtain Z class=

Sagot :

Answer:

z = -1 and 7/3

Step-by-step explanation:

after squaring each side you get:

z² + 8 = (1 - 2z)²

z² + 8 = 1 - 4z + 4z²

combine like terms to get this trinomial:

3z² - 4z - 7, which has factors of (3z - 7)(z + 1)

therefore, z = 7/3 and z = -1

The extraneous solutions for the Avery equation is [tex]z = \frac{7}{3}[/tex].

What is an equation?

The definition of an equation is a mathematical statement that shows that two mathematical expressions are equal. For instance, 3x + 5 = 14 is an equation, in which 3x + 5 and 14 are two expressions separated by an ‘equal’ sign. The most basic and simple algebraic equations consist of one or more variables in math.

Given equation

[tex]\sqrt{z^{2}+8 } =1-2z[/tex]

Squaring on both sides

[tex]z^{2} +8=(1-2z)^{2}[/tex]

[tex](1-2z)^{2}[/tex] is in the form of [tex](a-b)^{2} =a^{2} -2ab+b^{2}[/tex].

[tex]z^{2}+8=1-4z+4z^{2}[/tex]

[tex]4z^{2} -z^{2} -4z+1-8=0[/tex]

[tex]3z^{2} -4z-7=0[/tex]

[tex]3z^{2} +3z-7z-7=0[/tex]

[tex]3z(z+1)-7(z+1)=0[/tex]

[tex](3z-7)(z+1)=0[/tex]

[tex]z=\frac{7}{3}[/tex] or [tex]z=-1[/tex]

because [tex]\sqrt{z^{2}+8 } \ge 0[/tex] so [tex]1-2z \ge 0[/tex]

When [tex]z=\frac{7}{3}[/tex]          [tex]1-2 \times \frac{7}{3} =-\frac{11}{3} < 0[/tex](Abandon)

When z = -1          1 - 2 × (-1) = 1 + 2 = 3 > 0

So the extraneous solutions is [tex]z = \frac{7}{3}[/tex].

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