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babunello [35]
3 years ago
8

How do we check if a solution is extraneous? Use these equations as examples to help you explain:

Mathematics
1 answer:
Maurinko [17]3 years ago
5 0
Multiplying both sides of a rational equation by a variable expression introduces the possibility of extraneous solutions. Therefore, we must check the solutions against the set of restrictions. If a solution is a restriction, then it is not part of the domain and is extraneous.
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6x^2-9x +2=0 solve the quadratic equation using the quadratic formula
balu736 [363]

<u>ANSWER: </u>

The roots of 6 x^{2}-9 x+2=0 are \frac{9+\sqrt{33}}{4}, \frac{9-\sqrt{33}}{4}

<u>SOLUTION:</u>

Given, quadratic equation is 6 x^{2}-9 x+2=0 --- eqn (1)

We need to find the roots of given quadratic equation using quadratic formula.

Quadratic formula for general quadratic equation  a x^{2}+b x+c=0 is X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Here, for eqn (1) a = 6, b= -9 and c = 2

X=\frac{-(-9) \pm \sqrt{(-9)^{2}-4 \times6 \times 2}}{2 x 2}

X=\frac{9 \pm \sqrt{81-48}}{4}

\begin{array}{l}{X=\frac{9 \pm \sqrt{33}}{4}} \\ {X=\frac{9+\sqrt{33}}{4}, \frac{9-\sqrt{33}}{4}}\end{array}

Hence the roots of 6 x^{2}-9 x+2=0 are \frac{9+\sqrt{33}}{4}, \frac{9-\sqrt{33}}{4}

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What is the fill in the blank answers?
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