Question

How do you solve \[{b^2} = - 3b\] ?

Answer 1

Hint:To solve the given, combine all the like terms and shift the terms by this we can find the value of b as the equation is in the form of quadratic.

Complete step by step answer:
Let us write the given equation
\[{b^2} = - 3b\]
Shift the term \[ - 3b\] from RHS to the LHS and the equation we get is
\[{b^2} - \left( { - 3b} \right) = 0\]
Now let us combine and simplify all the like terms, we get
\[{b^2} + 3b = 0\]
We can see the obtained equation is of factored form, identify a, b and c using sum product pattern by finding two integers when multiplied together equal to c and when added together is equal to b i.e., The pair of integers we need to find for product is c and whose sum is b, in which here the product is 0 and sum is 3.
\[{b^2} + 3b + 0b = 0\]
Hence, the factors are
\[\left( {b + 3} \right)\left( {b + 0} \right) = 0\]
If any individual factor of the equation is equal to zero, them this implies that the entire expression
is equal to zero.
\[\left( {b + 3} \right) = 0\]
\[\left( {b + 0} \right) = 0\]
Now let us solve for the first factor i.e.,
\[\left( {b + 3} \right) = 0\]
Therefore, we get
\[b = - 3\]
Now let us solve for the second factor i.e.,
\[\left( {b + 0} \right) = 0\]
Therefore, we get
\[b = 0\]
Hence, the final solution is true for
\[b = - 3,0\]

Note: The key point to solve the given equation \[{b^2} = - 3b\] , shift the like terms and hence the equation is of the form \[{x^2} + bx + c\], in this given quadratic equation we need to find two integers whose product is equal to c and the sum is equal to b using AC method. Then solve each factor obtained by setting it to zero by this we can get the value of b of both the factors.