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Sagot :
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:
[tex]\textsf{D.}\quad5x^2+8x-12=2x^2-x[/tex]
Step-by-step explanation:
Quadratic Formula
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when}\:ax^2+bx+c=0[/tex]
A linear equation in the form [tex]y=mx+b[/tex] cannot be solved using the quadratic formula, as it is not a quadratic equation.
[tex]\textsf{A.}\quad 4x + 2 = 0[/tex]
This is a linear equation and therefore cannot be solved using the quadratic formula.
[tex]\begin{aligned}\textsf{B.}\quad 3x^2-4&=3x^2-4x\\\implies -4&=-4x\end{aligned}[/tex]
This is a linear equation and therefore cannot be solved using the quadratic formula.
[tex]\textsf{C.}\quad 2x=32[/tex]
This is a linear equation and therefore cannot be solved using the quadratic formula.
[tex]\textsf{D.}\quad5x^2+8x-12=2x^2-x[/tex]
[tex]\implies 3x^2+9x-12=0[/tex]
Therefore, this is a quadratic equation in the form [tex]ax^2+bx+c=0[/tex] and therefore can be solved quadratic formula.
a = 3, b = 9 and c = -12
Inputting these into the quadratic formula and solving for x:
[tex]\begin{aligned}\implies x &=\dfrac{-(9) \pm \sqrt{(9)^2-4(3)(-12)}}{2(3)}\\\\x& = \dfrac{-9 \pm \sqrt{225}}{6}\\\\x& = \dfrac{-9 \pm 15}{6}\\\\x&=\dfrac{6}{6},\dfrac{-24}{6}\\\\x&=1, -4\end{aligned}[/tex]
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