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3 years ago

15

Sam opened a savings account with a deposit of $50. Every month after that, he

1 answer:

Alexeev081 [22]3 years ago

7 0

Step-by-step explanation:

It is given that,

Sam opened a savings account with a deposit of $50. Every month after that, he deposited $20 more.

It is not a proportional relationship. As initial amount was $50 and every month he deposited $20 more. The ratio is not constant for the deposited amounts.

To make the relationship proportional, he must start with zero balance and then should have deposited $20 every month.

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Your friend Alex was solving the equation 2x^2+3x=20 and the work below. Tragically, she made an error in her work. First, expla

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

She has not made the Coefficient of x² as 1.

Therefore,

$x=\dfrac{-3+13}{4}=\dfrac{10}{4}=2.5\ OR\ x=\dfrac{-3-13}{4}=\dfrac{-16}{4}=-4$

Step-by-step explanation:

Alex was solving the equation

$2x^{2}+3x=20$

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

$2x^{2}+3x-20=0$

Step 2: Divide the whole by 2 we get

$x^{2}+\dfrac{3}{2}x-10=0$

Step 3 : Add 10 to both side

$x^{2}+\dfrac{3}{2}x-10+10=0+10\\\\x^{2}+\dfrac{3}{2}x=10$

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

$x^{2}+\dfrac{3}{2}x+\dfrac{9}{16}=10+\dfrac{9}{16}\\\\x^{2}+2\times \dfrac{3}{4}x+(\dfrac{3}{4})^{2}=\dfrac{169}{16}$

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

$(x+\dfrac{3}{4})^{2}=\dfrac{169}{16}$

Step 6 : Taking Square root

$(x+\dfrac{3}{4})=\pm \sqrt{\dfrac{169}{16}}=\pm \dfrac{13}{4}$

Step 7 : Subtract $\dfrac{3}{4}$ both side

$x+\dfrac{3}{4}-\dfrac{3}{4}=-\dfrac{3}{4}\pm \sqrt{\dfrac{169}{16}}=\pm \dfrac{13}{4}\\\\x=-\dfrac{3}{4} \pm \dfrac{13}{4}$

Step 8 : Finally we get

$x=\dfrac{-3+13}{4}=\dfrac{10}{4}=2.5\ OR\ x=\dfrac{-3-13}{4}=\dfrac{-16}{4}=-4$

Therefore,

$x=\dfrac{-3+13}{4}=\dfrac{10}{4}=2.5\ OR\ x=\dfrac{-3-13}{4}=\dfrac{-16}{4}=-4$

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