Question 28712: Find the value or values of p in the quadratic equation p^2+13p-30=0
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! Find the value or values of p in the quadratic equation p^2+13p-30=0
p^2+13p-30=0 ----(1)
p^2 + [(15)p+(-2)p] - 30 =0
(splitting the middle term as the sum of two terms in such a way that these two when multiplied(of course along with the signs)should give the product of the square term and the constant term.)
That is [13p =(15)p+(-2)p] and (15p)X(-2p) = -30p^2= (p^2)X(-30)]
p^2 + 15p-2p -30= 0
p(p+15)-2(p+15)=0
pt-2t=0 where t= p+15
t(p-2)=0
(p+15)(p-2)=0
(p+15) = 0 gives p= -15
(p-2)=0 gives p= 2
Answer: p= -15 and p =2
Verification: p= -15 in p^2+13p-30=0 ----(1)
LHS=(-15)^2 +13X(-15)-30 = 225-195-30=0 =RHs
And p= 2 in p^2+13p-30=0 ----(1)
LHS=(2)^2 +13X(2)-30 = 4+26-30= 0 =RHs
Therefore our values are correct.
|
|
|
