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]