SOLUTION: word problem for quadratic equation Brendan buys a block of shares of Laser Technology for $1875. When the share price increases by $4 per share, he sells all but 15 of them for

Question 106506: word problem for quadratic equation
Brendan buys a block of shares of Laser Technology for $1875. When the share price increases by $4 per share, he sells all but 15 of them for $1740. How many shares did he buy?
can anyone set-up the equation for this?? thanks for your help...

Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
Let x=number of shares he bought. p=share price he paid
A) xp=1875
B) (p+4)(x-15)=1740
.
A) x=1875/p
B) px+4x-15p-60=1740 Expand
p*1875/p + 4*1875/p - 15p-60=1740 substitute 1875/p for x
1875+7500/p-15p-60=1740 cancel
1875p+7500-15p^2-60p=1740p multiply both sides by p to eliminate fraction.
-15p^2+75p+7500=0
15p^2-75p-7500=0
15(p^2-5p-500)=0
15(p-25)(p+20)=0
p= -20 is extraneous to this problem.
p=$25
x=1875/25= 75 shares Answer
Check:
60*29=$1740.
.
Ed

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