Can someone help me with this? Thank you!

Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.

Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ; $x = \frac{2-(3)\sqrt{2}}{2}$

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the following condition must be followed:

Either (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

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

for the equation ax² + bx + c = 0

thus,

the roots are

$x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}$

or

$x = \frac{6\pm\sqrt{36-12}}{2}$

or

$x = \frac{6+\sqrt{24}}{2}$ and, x = $x = \frac{6-\sqrt{24}}{2}$

or

$x = \frac{6+2\sqrt{6}}{2}$ and, x = $x = \frac{6-2\sqrt{6}}{2}$

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

$x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}$

or

$x = \frac{4\pm\sqrt{16+56}}{4}$

or

$x = \frac{4+\sqrt{72}}{4}$ and, x = $x = \frac{4-\sqrt{72}}{4}$

or

$x = \frac{4+\sqrt{2^2\times3^2\times2}}{2}$ and, x = $x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}$

or

$x = \frac{4+(2\times3)\sqrt{2}}{2}$ and, x = $x = \frac{4-(2\times3)\sqrt{2}}{4}$

or

$x = \frac{2+3\sqrt{2}}{2}$ and, $x = \frac{2-(3)\sqrt{2}}{2}$

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ; $x = \frac{2-(3)\sqrt{2}}{2}$

7 0

2 years ago

Divide using long division. HELP ASAP!!

pav-90 [236]

Answer:

The answer to your question is the first option

Step-by-step explanation:

Write the division

7x³ - 7x² + 6x +4 Result

x + 1 7x⁴ - 0x³ - 1x² + 10x + 15

-7x⁴ - 7x³

0 - 7x³ - 1x²

+ 7x³ +7x²

0 + 6x² + 10x

- 6x² - 6x

0 + 4x + 15

- 4x - 4

0 + 11 Remainder

Result = 7x³ - 7x² + 6x + 4 + 11 / x + 1

8 0

3 years ago

If i= the square root of -1 what is the value of i^3

BARSIC [14]

i =√-1

i^2 = -1

so

i^3

= i^2 * i

= -1√-1

= - √-1

5 0

2 years ago

Pls help ill give u brainliest

Monica [59]

Answer:

Answers:

Rate of change = -5

Initial value = 25

Step-by-step explanation:

Each time x increases by 2 (eg from x = 1 to x = 3), the value of y drops by 10 (eg from y = 20 to y = 10)

Therefore the slope is...

slope = rise/run = (change in y)/(change in x) = -10/2 = -5

slope = -5

So each time x increases by 1, y will decrease by 5

Flip things around: each time x decreases by 1, y will increase by 5

So the pair (x,y) = (1,20) shown in row 1 of the table leads to (0,25) based on the rule above. Another way to see this is to plug m = -5, x = 1 and y = 20 into y = mx+b and solve for b to get b = 25

8 0

2 years ago

In the equation (m^5/6) (m^1/6)^7 = m^x , what is the value of x that makes the equation true?

vlabodo [156]

${m}^{ \frac{5}{6} } \times (m {}^{ \frac{1}{6} } ) {}^{7 } = m {}^{x}$

$m {}^{ \frac{5}{6} } \times m {}^{ \frac{7}{6} } = m {}^{x}$

${m}^{ \frac{5 \times 7}{6 \times 6} } = {m}^{x}$

$\frac{35}{36} = x$

6 0

2 years ago