Question

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after being rearranged and simplified, which equation could be solved using the quadratic formula


A. 4x+2=0


B. 3x2-4=3x^2-4x


C. 2x=32


D. 5x^2+8x-12=2x2-x

Answer 1

Option D

• 5x²+8x-12=2x²-x
• 5x²-2x²+8x+x-12=0
• 3x²+9x-12=0
• x²+3x-4=0

Now we can apply quadratic formula to find two zeros

Answer 2

`\textsf{D.}\quad5x^2+8x-12=2x^2-x`

Explanation:

Quadratic Formula

`x=(-b \pm √(b^2-4ac) )/(2a)\quad\textsf{when}\:ax^2+bx+c=0`

A linear equation in the form
`y=mx+b` cannot be solved using the quadratic formula, as it is not a quadratic equation.

`\textsf{A.}\quad 4x + 2 = 0`

This is a linear equation and therefore cannot be solved using the quadratic formula.

`\begin{aligned}\textsf{B.}\quad 3x^2-4&=3x^2-4x\\\implies -4&=-4x\end{aligned}`

This is a linear equation and therefore cannot be solved using the quadratic formula.

`\textsf{C.}\quad 2x=32`

This is a linear equation and therefore cannot be solved using the quadratic formula.

`\textsf{D.}\quad5x^2+8x-12=2x^2-x`

`\implies 3x^2+9x-12=0`

Therefore, this is a quadratic equation in the form
`ax^2+bx+c=0` and therefore can be solved quadratic formula.

a = 3, b = 9 and c = -12

Inputting these into the quadratic formula and solving for x:

`\begin{aligned}\implies x &=(-(9) \pm √((9)^2-4(3)(-12)))/(2(3))\\\\x& = (-9 \pm √(225))/(6)\\\\x& = (-9 \pm 15)/(6)\\\\x&=(6)/(6),(-24)/(6)\\\\x&=1, -4\end{aligned}`