Help would be helpful ;)

Question

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Let p and q be the two distinct solutions to the equation $(x-3)(x+3)=21x-63$ . If $p>q$ , what is the value of $p-q$ ?

Hint: Does not equal 7

Keihaku Nov 28, 2022

Guest · Answer 1 · 2022-11-28T06:16:30

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The roots are 3 and 5, so p - q = 5 - 3 = 2.

Guest Nov 28, 2022

+660

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That's incorrect, sorry; can someone else pls help

Keihaku Nov 28, 2022

edited by Keihaku Nov 28, 2022

Guest · Answer 2 · 2022-11-28T07:09:57

The quadratic arranges to $x^2 - 15x + 36 = 0$

Then by the quadratic formula,

$x = \frac{15 \pm \sqrt{15^2 - 4 \cdot 36}}{2} = \frac{15 \pm 9}{2} = 12, 3$

Therefore, $p - q = 12 - 3 = \boxed{9}$

geno3141 · Answer 3 · 2022-11-29T01:03:14

(x - 3)(x + 3) = 21x - 63

x² - 9 = 21x - 63

x² - 21x + 54 = 0

p = [ 21 + sqrt( 21² - 4·1·54 ) ] / 2 = [21 + sqrt( 225 ) ] / 2 = [ 21 + 15 ] / 2 = 18

q = [ 21 - sqrt( 21² - 4·1·54 ) ] / 2 = [21 - sqrt( 225 ) ] / 2 = [ 21 - 15 ] / 2 = 3

p - q = 18 - 3 = 15