3 years ago

A quadratic equation x^2-8x+9=0 is rewritten in the form (x-p)^2=q. What are the values of p and q?

Mathematics

1 answer:

grin007 [14]3 years ago

8 0

X^2-(2)(x)(4)+16-7=0
x^2-(2)(x)(4)+4^2 =7
(x-4)^2=7

So value of p is 4 and q is 7

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-(m - 6) = 3m + 14<br> Multiply m- 6 by -1

Angelina_Jolie [31]

For this case we have the following equation:

$- (m-6) = 3m + 14$

We apply distributive property on the left side of the equation, taking into account that:

$- * + = -\\- * - = +$

So:

$-m + 6 = 3m + 14$

Subtracting 3m from both sides of the equation:

$-m-3m + 6 = 14$

Equal signs are added and the same sign is placed:

$-4m + 6 = 14$

We subtract 6 from both sides of the equation:

$-4m = 14-6\\-4m = 8$

We divide between -4 to both sides of the equation:

$m = \frac {8} {- 4}\\m = -2$

Thus, the solution of the equation is $m = -2$

ANswer:

$m = -2$

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