Respuesta :

carlosego

For this case we must indicate which of the equations shown can be solved using the quadratic formula.

By definition, the quadratic formula is applied to equations of the second degree, of the form:

[tex]ax ^ 2+ bx+ c = 0[/tex]

Option A:

[tex]2x ^ 2-3x +10 = 2x + 21[/tex]

Rewriting we have:

[tex]2x ^ 2-3x-2x+ 10-21 = 0\\2x ^ 2-5x-11 = 0[/tex]

This equation can be solved using the quadratic formula

Option B:

[tex]2x ^ 2-6x-7 = 2x ^ 2[/tex]

Rewriting we have:

[tex]2x ^ 2-2x ^ 2-6x-7 = 0\\-6x-7 = 0[/tex]

It can not be solved with the quadratic formula.

Option C:

[tex]5x ^ 2 + 2x-4 = 2x ^ 2[/tex]

Rewriting we have:

[tex]5x ^ 2-2x ^ 2 + 2x-4 = 0\\3x ^ 2 + 2x-4 = 0[/tex]

This equation can be solved using the quadratic formula

Option D:

[tex]5x ^ 3-3x + 10 = 2x ^ 2[/tex]

Rewriting we have:

[tex]5x ^ 3-2x ^ 2-3x + 10 = 0[/tex]

It can not be solved with the quadratic formula.

Answer:

A and C

alinakincsem

Answer:

The correct answer options are:

A) [tex]2x^{2} -3x+10=2x+21[/tex]

C) [tex]5x^2+2x-4=2x^2[/tex]

Step-by-step explanation:

Quadratic formula can be used to solve the equations in the form [tex]ax^2+bx+c[/tex]. Re-writing the equations to see if they can be solved using the quadratic formula.

A) [tex]2x^{2} -3x+10=2x+21[/tex]:

[tex]2x^2+3x-2x+10-21=0[/tex]

[tex]2x^{2} -5x-11=0[/tex]


B) [tex]2x^2-6x-7=2x^2[/tex]

[tex]2x^{2} -2x^{2} -6x-7=0[/tex]

[tex]-6x-7=0[/tex]


C) [tex]5x^2+2x-4=2x^2[/tex]

[tex]5x^{2} -2x^2+2x-4=0[/tex]

[tex]3x^{2} +2x-4=0[/tex]


D) [tex]5x^3-3x+10=2x^2[/tex]

[tex]5x^3-2x^2-3x+10=0[/tex]


Therefore, option A) [tex]2x^{2} -3x+10=2x+21[/tex] and option C) [tex]5x^2+2x-4=2x^2[/tex] can be solved using the quadratic formula.

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