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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

 Nov 28, 2022
 #1
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The roots are 3 and 5, so p - q = 5 - 3 = 2.

 Nov 28, 2022
 #2
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That's incorrect, sorry; can someone else pls help

Keihaku  Nov 28, 2022
edited by Keihaku  Nov 28, 2022
 #3
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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, \(p - q = 12 - 3 = \boxed{9}\)

 Nov 28, 2022
 #4
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also incorrect

Keihaku  Nov 28, 2022
 #5
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+2

(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

 Nov 29, 2022

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