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1 vote
How do u doo thiss?? :)​

How do u doo thiss?? :)​-example-1

2 Answers

4 votes

Answer: a = 5

b = -33

Explanation:

They want you to put the quadratic in vertex form. You need to complete the square to do this.

x² + 10x - 8 >Make space to complete square

x² + 10x - 8 >Take middle term 10, divide by 2 then square

that
((10)/(2)) ^(2) = 25

Use +25 to fill in to complete the square but you

can't just add 25 to an equation, it changes it.

You need to counterbalance it by also

subtract 25. so 25-25= 0 you are not changing

x² + 10x +25 - 8 -25 >Combine the -8 and -25

x² + 10x +25 - 33 >Factor the first 3 terms x² + 10x +25

(x+5)(x+5) -33 >(x+5)(x+5) = (x+5)²

(x+5)² - 33 >this is the form they wanted.

a = 5

b = -33

User Clarus Dignus
by
8.5k points
2 votes


\sf (x + a)^2 + b = x^2 + 10x - 8 \\

Comparing coefficients, we can equate the corresponding terms:


\sf 2ax = 10x \\ (coefficient of x terms)


\sf a^2 + b = -8 \\ (constant term)

From the first equation, we can determine the value of
\sf a \\:


\sf 2ax = 10x \\


\sf 2a = 10 \\ (canceling out x from both sides)


\sf a = (10)/(2) \\

Simplifying further:


\sf a = 5 \\

Now, we can substitute the value of
\sf a \\ into the second equation:


\sf a^2 + b = -8 \\


\sf 5^2 + b = -8 \\


\sf 25 + b = -8

Solving for
\sf b \\:


\sf b = -8 - 25 \\


\sf b = -33 \\

Therefore, the values of
\sf a \\ and
\sf b \\ are
\sf 5 \\ and
\sf -33 \\, respectively.

Thus, the expression
\sf x^2 + 10x - 8 \\ can be written in the form
\sf (x + 5)^2 - 33 \\.


\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}

♥️
\large{\textcolor{red}{\underline{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}

User Rob Tanzola
by
8.7k points

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