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3 years ago

2xsquare-6x+3=0 by completing square method

Mathematics

1 answer:

Arturiano [62]3 years ago

5 0

Answer:

x = 2.36603 or 0.633975

Step-by-step explanation:

Here we want to solve the quadratic equation using the completing the square method.

2x^2-6x + 3 = 0

divide through by 2 first;

x^2 -3x + 3/2 = 0

x*2 -3x = -3/2

Half the x term will be -3/2 and squaring it is 9/4

we add this to both sides

x^2-3x/2 + 9/4 = -3/2 + 9/4

(x-3/2)^2 = 3/4

Finding square root of both sides

x-3/2 = √(3/4)

x = 3/2 ± √(3/4)

x = 3/2 + √(3/4) or x = 3/2-√(3/4)

which becomes

x = 2.36603 or 0.633975

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Math question #1 please show steps

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Step-by-step explanation:

An approximation of an integral is given by:

$\displaystyle \int_a^bf(x)\, dx\approx \sum_{k=1}^nf(x_k)\Delta x\text{ where } \Delta x=\frac{b-a}{n}$

First, find Δx. Our a = 2 and b = 8:

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The left endpoint is modeled with:

$x_k=a+\Delta x(k-1)$

And the right endpoint is modeled with:

$x_k=a+\Delta xk$

Since we are using a Left Riemann Sum, we will use the first equation.

Our function is:

$f(x)=\cos(x^2)$

Therefore:

$f(x_k)=\cos((a+\Delta x(k-1))^2)$

By substitution:

$\displaystyle f(x_k)=\cos((2+\frac{6}{n}(k-1))^2)$

Putting it all together:

$\displaystyle \int_2^8\cos(x^2)\, dx\approx \sum_{k=1}^{n}\Big(\cos((2+\frac{6}{n}(k-1))^2)\Big)\frac{6}{n}$

Thus, our answer is C.

*Note: Not sure why they placed the exponent outside the cosine. Perhaps it was a typo. But C will most likely be the correct answer regardless.

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