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Q. 2

Expert-verifiedFound in: Page 653

Book edition
6th

Author(s)
Sullivan

Pages
1200 pages

ISBN
9780321795465

To complete the square of ${x}^{2}-3x$ , add$\_\_\_\_\_\_$ .

To complete the square of ${x}^{2}-3x$, add role="math" localid="1646794628543" $\frac{\mathbf{9}}{\mathbf{4}}$.

Given Expression:

${x}^{2}-3x$

We want to complete the square of the given expression with the missing terms.

What it actually means to complete a square is to find a number $a$ that when added to the given expression enables the whole expression, including the added number, to be written as a square like in the following example:

${x}^{2}-3x+a={(x+b)}^{2}$

First expand the right side of the equality ${x}^{2}+bx+{b}^{2}$.

Both sides of the equality are polynomials and even more, they are equal.

Therefore, coefficients in front of the same power of the variable $x$ must be the same.

For role="math" localid="1646795833717" ${x}^{2}:1=1$ the quality obviously holds. Moving on to $x:-3=2b$

Therefore $b=-\frac{3}{2}$. And finally $a={b}^{2}={(-\frac{3}{2})}^{2}=\frac{9}{4}$

Therefore, in order to complete the square add $\frac{9}{4}$ to the expression. You can check if the answer is correct by finding the square: ${x}^{2}-3x+\frac{9}{4}={(x-\frac{3}{2})}^{2}$

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