Your friend Alex was solving the equation 2x^2+3x=20 and the work below. Tragically, she made an error in her work. First, explain her error (and why), then fix her mistake by redoing the problem correctly

Respuesta :

Answer:

Her error is that For solving by Completing the square method.

She has not made the Coefficient of x² as 1.

Therefore,

[tex]x=\dfrac{-3+13}{4}=\dfrac{10}{4}=2.5\ OR\ x=\dfrac{-3-13}{4}=\dfrac{-16}{4}=-4[/tex]

Step-by-step explanation:

Alex was solving the equation

[tex]2x^{2}+3x=20[/tex]

She is solving for x by Completing the square method

Her error is that For solving by Completing the square method

First step is to make Coefficient of x² as 1 ( one )

She has not made the Coefficient of x² as 1

The Require steps for the equation by Completing the square method are

Step 1 : Given

[tex]2x^{2}+3x-20=0[/tex]

Step 2: Divide the whole by 2 we get

[tex]x^{2}+\dfrac{3}{2}x-10=0[/tex]

Step 3 : Add 10 to both side

[tex]x^{2}+\dfrac{3}{2}x-10+10=0+10\\\\x^{2}+\dfrac{3}{2}x=10[/tex]

Step 4 : Add [tex](\dfrac{1}{2}coefficient\ of\ x)^{2}=\dfrac{9}{16}[/tex] to both the side.

[tex]x^{2}+\dfrac{3}{2}x+\dfrac{9}{16}=10+\dfrac{9}{16}\\\\x^{2}+2\times \dfrac{3}{4}x+(\dfrac{3}{4})^{2}=\dfrac{169}{16}[/tex]

Step 5 : We have A² + 2AB + B² = (A+B)²

[tex](x+\dfrac{3}{4})^{2}=\dfrac{169}{16}[/tex]

Step 6 : Taking Square root

[tex](x+\dfrac{3}{4})=\pm \sqrt{\dfrac{169}{16}}=\pm \dfrac{13}{4}[/tex]

Step 7 : Subtract [tex]\dfrac{3}{4}[/tex] both side

[tex]x+\dfrac{3}{4}-\dfrac{3}{4}=-\dfrac{3}{4}\pm \sqrt{\dfrac{169}{16}}=\pm \dfrac{13}{4}\\\\x=-\dfrac{3}{4} \pm \dfrac{13}{4}[/tex]

Step 8 : Finally we get

[tex]x=\dfrac{-3+13}{4}=\dfrac{10}{4}=2.5\ OR\ x=\dfrac{-3-13}{4}=\dfrac{-16}{4}=-4[/tex]

Therefore,

[tex]x=\dfrac{-3+13}{4}=\dfrac{10}{4}=2.5\ OR\ x=\dfrac{-3-13}{4}=\dfrac{-16}{4}=-4[/tex]

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