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3 years ago

15

How to solve this equation 6x^2-2x+5+2x^2-7x-11

1 answer:

postnew [5]3 years ago

6 0

x= $\frac{9+\sqrt{273} }{16}$

OR

x= $\frac{9-\sqrt{273} }{16}$

Step-by-step explanation:

The given equation is

$6x^{2}$ -2x+5+ $2x^{2}$ -7x-11

<u>Step 1: Arrange the equation</u>

$6x^{2}$ + $2x^{2}$ -2x-7x-11+5

<u>Step 2: Simplify the equation</u>

$8x^{2}$ -9x-6

<u>Step 3: Identify a,b,c in the equation</u>

a=8

b=-9

c=-6

<u>Step 4: Apply quadratic formula to find the possible values of x</u>

x = $\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

x= $\frac{-(-9)±\sqrt{(-9)^{2}-4(8)(-6) } }{2(8)}$

x= $\frac{9+\sqrt{273} }{16}$

OR

x= $\frac{9-\sqrt{273} }{16}$

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Select the polynomial that is a perfect square trinomial. 36x2 − 4x 16 16x2 − 8x 36 25x2 9x 4 4x2 20x 25.

Naya [18.7K]

The polynomial a,b,c,are not perfect square polynomial and the polynomial d is perfect square polynomial.

The given polynomial is

$a) 36x^2-4x+16$

What is the form of perfect square polynomial?

$(ax+b)^2$

we solve this method by using perfect square method

add and subtract 1/9

$36x^2-4x+16+\frac{1}{9}-\frac{1}{9}$

factor 36

$\frac{143}{9}+36(x^2-\frac{x}{9}+\frac{1}{324} )$

Now complete the square

$\frac{143}{9}+36(x^2-\frac{1}{18})^2$

Therefore this is not perfect square trinomial.

Similarly for

$b) 16x^2-8x+36$

Complete square is,

$16(x-\frac{1}{4})^2+35$

This polynomial is also not perfect square trinomial.

$c) 25x^2+9x+4\\$

complete square is,

$25(x+\frac{9}{50})^2+\frac{319}{100}$

This polynomial is not perfect square trinomial.

$d)4x^2+20x+25\\$

complete square is,

$(2x+5)^2$

This polynomial is perfect square trinomial.

Therefore,

The polynomial a,b,c,are not perfect square polynomial and the polynomial d is perfect square polynomial.

To learn more about perfect square trinomial visit:

brainly.com/question/1538726

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2 years ago

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the corre

vovikov84 [41]

Answer:

Step-by-step explanation:

$b^{2*3} =b^{6}$

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3 years ago

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ivolga24 [154]

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Step-by-step explanation:

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Assume that the terminal side of thetaθ passes through the point (negative 12 comma 5 )(−12,5) and find the values of trigonomet

zmey [24]

Answer:

$\sin \theta = \dfrac{5}{13}$ and $\sec \theta = -\dfrac{13}{12}$

Step-by-step explanation:

Assume that the terminal side of thetaθ passes through the point (−12,5).

In ordered pair (-12,5), x-intercept is negative and y-intercept is positive. It means the point lies in 2nd quadrant.

Using Pythagoras theorem:

$hypotenuse^2=perpendicular^2+base^2$

$hypotenuse^2=(5)^2+(12)^2$

$hypotenuse^2=25+144$

$hypotenuse^2=169$

Taking square root on both sides.

$hypotenuse=13$

In a right angled triangle

$\sin \theta = \dfrac{opposite}{hypotenuse}$

$\sin \theta = \dfrac{5}{13}$

$\sec \theta = \dfrac{hypotenuse}{adjacent}$

$\sec \theta = \dfrac{13}{12}$

In second quadrant only sine and cosecant are positive.

$\sin \theta = \dfrac{5}{13}$ and $\sec \theta = -\dfrac{13}{12}$

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4 years ago

Answer and explain. 30 points given out

Naddik [55]

This is what I get

(120-60%)-50%=24

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3 years ago