Using the Completing the Square method, what are the zeros of the quadratic function f(x) = 2x2 + 8x - 3?

Mathematics

1 answer:

Ipatiy [6.2K]1 year ago

4 0

Answer:

Step-by-step explanation:

1) to rewrite the given equation:

f(x)=2(x²+4x)-3; ⇔ f(x)=2(x²+4x+4)-3-8; ⇔ f(x)=2(x²+4x+4)-11; ⇔ f(x)=2(x+2)²-11;

2) the roots are:

$\left[\begin{array}{ccc}x=-\sqrt{\frac{11}{2}}-2 \\x=\sqrt{\frac{11}{2}}-2 \end{array}$

3) finally, the correct answer is A.

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How do you this question (pre-cal)

Novay_Z [31]

The goal to proving identities is to transform one side into the other. We can only pick one side to transform while the other side stays the same the entire time. The general rule of thumb is to transform the more complicated side (though there may be exceptions to this guideline).

So I'll take the left hand side and try to turn it into $\csc^2( B )$

One way we can do that is through the following steps:

$\frac{\tan(B) + \cot(B)}{\tan(B)} = \csc^2(B)\\\\\frac{\tan(B)}{\tan(B)} + \frac{\cot(B)}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\frac{1}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\cot(B) = \csc^2(B)\\\\1 + \cot^2(B) = \csc^2(B)\\\\1 + \frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)}{\sin^2(B)}+\frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)+cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{1}{\sin^2(B)} = \csc^2(B)\\\\\csc^2(B)=\csc^2(B) \ \ {\Large \checkmark}\\\\$