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3 years ago

Write an equation for each problem. The square of a number is 8 more than twice the number.

Mathematics

1 answer:

avanturin [10]3 years ago

3 0

Answer:

The equation is equal to

$x^{2}=2x+8$

Step-by-step explanation:

Let

x -----> the number

we know that

The equation that represented the problem is equal to

$x^{2}=2x+8$

$x^{2}-2x-8=0$

Solve the quadratic equation by graphing

The solution are x=-2 and x=4

see the attached figure

therefore

The number can be -2 or 4

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Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

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Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

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Angle a = 126. What is the measure of angle b? Explain how you calculated your answer.

Angle a = 126 Write an equation(s) in terms of b to find the measure of angle h.

Calculate the measure of angle h, using the equation(s) you wrote for Part B.

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2021 Muminate Education Inc Your input: factor x2+4x+3.

To factor the quadratic function x2+4x+3, we should solve the corresponding quadratic equation x2+4x+3=0.

Indeed, if x1 and x2 are the roots of the quadratic equation ax2+bx+c=0, then ax2+bx+c=a(x−x1)(x−x2).

Solve the quadratic equation x2+4x+3=0.

The roots are x1=−1, x2=−3 (use the quadratic equation calculator to see the steps).

Therefore, x2+4x+3=1(x+1)(x+3).

(x2+4x+3)=1(x+1)(x+3)

Step-by-step explanation:

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