x^{2}+4 x -192
x^{2}+4 x -192Step By Step Solution :
Let’s assume
p(x)=x^{2}+4x-192For zeroes of p(x)=0, we have to use factorization :
p(x)=x^{2}+16x-12x-192\implies x \times(x+16)-12 \times(x+16)
\implies (x-12) \cdot(x+16)=p(x)=0
For p(x)=0,
Either
x-12=0\\ \therefore x=12
Or
x+16=0 \\ \therefore x=-16
Hence, the roots of the polynomial :
x_{1}=-16 \\\\ x_{2}=12If
x_{1},x_{2} are roots of the polynomial
ax^{2}+bx+c=0then, the relation :
Sum \ of \ roots = x_{1}+x_{2} = \frac{-b}{a}=\frac {Coefficient\ of \ x}{Coeffficient \ of \ x^{2}}Product \ of \ roots =x_{1} \cdot x_{2}= \frac{c}{a}=\frac {Constant\ Term}{Coeffficient \ of\ x^{2}}If -16 and 12 are zeroes of
x^{2}+4x-192then,
Sum \ of zeros =x_{1}+x_{2}=-16+12=\frac{-4}{1}After subtract in left hand side:
-4=-4
Left hand side and right hand side are equal.
Now,
Product of zeroes=
-16\times 12=\frac{-192}{1}After multiply in left hand side:
-192=-192
Left hand side and right hand side are equal.
Solution ::The zeroes of polynomial
x^{2}+4x-192are -16 and 12:
Now,
sum of zeroes =-16+12=-4= \frac{-4}{1}=\frac{-~Coefficient~ of ~x}{Coefficient ~of~ x^{2}}Again,
Product of zeroes =-16\times 12=-192=\frac{-192}{1}=\frac{Constant ~term}{Coefficient ~of~ x^{2}}.Similar Post
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