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2 years ago

Multiply 7x(x^- x+2)

Mathematics

1 answer:

Pie2 years ago

6 0

Rewrite (7x−2) 2 (7x-2) 2 as (7x−2) (7x−2) (7x-2) (7x-2).

(7x−2) (7x−2) (7x-2) (7x-2)

Expand (7x−2)(7x−2)(7x-2)(7x-2) using the FOIL Method.

Apply the distributive property.

7x (7x−2) −2 (7x−2) 7x (7x-2)-2 (7x-2)

Apply the distributive property.

7x(7x)+7x⋅−2−2(7x−2)7x (7x)+7x⋅-2-2(7x-2)

Apply the distributive property.

7x(7x)+7x⋅−2−2(7x)−2⋅−27x(7x)+7x⋅-2-2(7x)-2⋅-2

Simplify and combine like terms.

Simplify each term.

Multiply xx by xx.

7⋅7x2+7x⋅−2−2(7x)−2⋅−27⋅7x2+7x⋅-2-2(7x)-2⋅-2

Multiply 77 by 77.

49x2+7x⋅−2−2(7x)−2⋅−249x2+7x⋅-2-2(7x)-2⋅-2

Multiply −2-2 by 77.

49x2−14x−2(7x)−2⋅−249x2-14x-2(7x)-2⋅-2

Multiply 77 by −2-2.

49x2−14x−14x−2⋅−249x2-14x-14x-2⋅-2

Multiply −2-2 by −2-2.

49x2−14x−14x+449x2-14x-14x+4

Subtract 14x14x from −14x-14x.

49x2−28x+4

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3 years ago

3(x – 2) + 4x = 10(x + 1)

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3 years ago

Read 2 more answers

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3 years ago

Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat

zavuch27 [327]

Answer:

The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term $x^2$ , which means that its graph shows branches going up. Therefore:

1) if its vertex is ON the x axis, there would be one solution (root) to the equation.

2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:

We recall that the x-position of the vertex for a quadratic function of the form $f(x)=ax^2+bx+c$ is given by the expression: $x_v=\frac{-b}{2a}$

Since in our case $a=1$ and $b=5$ , we get that the x-position of the vertex is: $x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

$y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}$

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the $x_v=-\frac{5}{2}$ . Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.

When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):

$f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0$

The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.

Then, the two roots are: x = -1 and x = -4.

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3 years ago