Question

Find the sum of and product of zeroes of \[{{x}^{2}}=12-x\].

Answer 1

Hint: In this problem, we have to find the sum and the product of the roots of the given equation. We know that in a general equation \[a{{x}^{2}}+bx+c=0\], the sum of the roots is \[-\dfrac{b}{a}\] and the product of the roots is \[\dfrac{c}{a}\], we can now compare the given equation and the general equation and find the value of a, b, c and substitute in the required formula to get the required result.

Complete step by step answer:
Here we have to find the sum and the product of the roots of the given equation.
We know that the given equation is,
 \[{{x}^{2}}=12-x\]
 We can now write it as,
\[\Rightarrow {{x}^{2}}+x-12=0\]…… (1)
We know that the general equation is,
 \[\Rightarrow a{{x}^{2}}+bx+c=0\]…….. (2)
 We can now compare the equations (1) and (2), we get
a = 1, b = 1, c = -12.
We know that,
The sum of the roots = \[-\dfrac{b}{a}\]
We can now substitute the value of b and a, we get
The sum of the roots = \[-\dfrac{1}{1}=-1\]
The product of the roots =\[\dfrac{c}{a}\]= \[\dfrac{-12}{1}=-12\].
Therefore, the sum of the roots is -1 and the product of the roots is -12.

Note: We should always remember that for the general equation \[a{{x}^{2}}+bx+c=0\], the sum of the roots is \[-\dfrac{b}{a}\] and the product of the roots is \[\dfrac{c}{a}\], we can now compare the given equation and the general equation and find the value of a, b, c.