Question

How do you factor and solve \[{{x}^{2}}+3x=28\]?

Answer 1

Hint: For the given problem we have to factorize and solve the equation\[{{x}^{2}}+3x=28\]. First of all we have to factorize the given equation and then we have to find the solution of the equation. For factorization of the equation transfer the term 28 from RHS (Right hand side) to LHS (Left hand side) and then factorize and find the solution of the equation.

Complete step by step answer:
For the given question we are given to factor and solve the equation\[{{x}^{2}}+3x=28\].
Let us consider
\[{{x}^{2}}+3x=28.........\left( 1 \right)\]
To solve the equation (1) we have to factorize the equation (1) and then we will get our solution easily.
Let us write the term 3x as 7x-4x in equation (1), we get
\[\Rightarrow {{x}^{2}}+7x-4x=28\]
Let us consider the above equation as equation (2).
\[{{x}^{2}}+7x-4x=28.............\left( 2 \right)\]
Transferring 28 from LHS (left hand side to RHS (Right hand side).
\[\Rightarrow {{x}^{2}}+7x-4x-28=0\]
Let us consider the above equation as equation (3)
\[{{x}^{2}}+7x-4x-28=0.......\left( 3 \right)\]
Rewriting the equation as
\[\Rightarrow {{x}^{2}}-4x+7x-28=0\]
Let us consider
\[{{x}^{2}}-4x+7x-28=0.........\left( 4 \right)\]
Taking x common from the first two terms and 7 common from the next two terms from the equation (4).
\[\Rightarrow x\left( x-4 \right)+7\left( x-4 \right)=0\]
By simplifying this we get-
\[\Rightarrow \left( x+7 \right)\left( x-4 \right)=0\]
Let us consider the above equation as equation (5).
Therefore values of x from the equation (5) are
\[\begin{align}
  & \Rightarrow x=4\text{ } \\
 & \Rightarrow \text{x=-7} \\
\end{align}\]
Let us consider the values of x are\[{{X}_{1}}\] and\[{{X}_{2}}\].
\[\begin{align}
  & {{X}_{1}}=4 \\
 & {{X}_{2}}=-7 \\
\end{align}\]

Therefore the solutions for equation (1) are \[{{X}_{1}}\] and\[{{X}_{2}}\].

Note: The alternate method for solving this problem is direct division. While splitting the term \[3x\] if we split with different factors of 28 then we cannot factorize our equation. To check whether the solutions are correct or not we have to substitute the ‘x’ values in the equation, then if we have 0 as a solution then our ‘x’ values will be correct.