SOLUTION: The quadratic function has 2 real zeroes that differ by 18. Find the value of c. f(x)= 2x^2 -24x+c I've tried to plug in numbers but I dont get answers that differs by 18

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Question 1081768: The quadratic function has 2 real zeroes that differ by 18. Find the value of c.
f(x)= 2x^2 -24x+c
I've tried to plug in numbers but I dont get answers that differs by 18

Found 3 solutions by Boreal, ikleyn, MathTherapy:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
Quadratic formula is
x=(1/4)(-24+/- sqrt (576-8c))
The value +/- has to be 36, because divided by 4,it will be +/-9, and that will make the two roots 18 apart.
Therefore, sqrt (576-8c)=36
576-8c=1296
-8c=720
c=-90 ANSWER
roots are 15 and -3.







Answer by ikleyn(52847)   (Show Source): You can put this solution on YOUR website!
.
Let p and q are the roots.

Then, according to Vieta's theorem,

p + q =  = 12.

The second condition is 

p - q = 18.


So, you have this system of equations

p + q = 12,
p - q = 18.

Add the equations. You will get

2p = 12 + 18 = 30.

Hence, p = 15.

Then q = 12 - p = 12 - 15 = -3.


Thus the roots are 15 and -3.

Then, according to Vieta's theorem,  = 15*(-3) = -45.

Then c = 2*(-45) = -90.


Answer.  c = -90, and the roots are 15 and -3.


Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!

The quadratic function has 2 real zeroes that differ by 18. Find the value of c.
f(x)= 2x^2 -24x+c
I've tried to plug in numbers but I dont get answers that differs by 18
Let zeroes (roots) be 
Then: =====> =====> ------ eq (i)
Also, ----- Given ------ eq (ii)
------ Adding eqs (ii) & (i)

------- Substituting 15 for in eq (i)

Also, ======>


--------- Cross-multiplying

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