Question

How do you solve \[\left( 3x-5 \right)\left( 7x+8 \right)=0\]?

Answer 1

Hint: This type of problem is based on the concept of factoring and solving x. First, we have to consider the given equation. Since the given equation is the product of two functions of x, these are the factor of the equation. Given, the equation is equal to 0, and hence the factors are also equal to 0. Consider 3x-5=0. Add 5 on both the sides of the equation and divide the equation by 3 to find the value of x. Now, we have to consider 7x+8=0. Subtract 8 from both the sides of the equation and divide the obtained equation by 7. Thus, we get the required answer.

Complete step by step answer:
According to the question, we are asked to solve \[\left( 3x-5 \right)\left( 7x+8 \right)=0\].
 We have been given the equation is \[\left( 3x-5 \right)\left( 7x+8 \right)=0\]. ---------(1)
The given equation is the product of two functions with variable x.
There are the factors of the equation (1).
Therefore, the factors are \[\left( 3x-5 \right)\] and \[\left( 7x+8 \right)\].
Since the equation is equal to 0, the factors will also be equal to 0.
Therefore, we get
\[3x-5=0\] -----------(2)
\[7x+8=0\] -----------(3)
We have to now find the values of x.
First, consider the equation (2).
Add 5 on both the sides of the equation.
\[\Rightarrow 3x-5+5=0+5\]
\[\Rightarrow 3x-5+5=5\]
We know that terms with the same magnitude and opposite signs cancel out.
\[\Rightarrow 3x=5\]
Now, we have to divide the whole equation by 3.
\[\Rightarrow \dfrac{3x}{3}=\dfrac{5}{3}\]
We find that 3 are common in both the numerator and denominator of the LHS. On cancelling 3, we get
\[x=\dfrac{5}{3}\]
Now, consider equation (3).
Subtract the whole equation by 8.
\[\Rightarrow 7x+8-8=0-8\]
\[\Rightarrow 7x+8-8=-8\]
We know that terms with the same magnitude and opposite signs cancel out.
\[7x=-8\]
Now, we have to divide the whole equation by 7.
\[\Rightarrow \dfrac{7x}{7}=\dfrac{-8}{7}\]
We find that 7 are common in both the numerator and denominator of the LHS. On cancelling 7, we get
\[x=\dfrac{-8}{7}\]

Therefore, the values of x in the equation \[\left( 3x-5 \right)\left( 7x+8 \right)=0\] are \[\dfrac{5}{3}\] and \[\dfrac{-8}{7}\].

Note: Whenever we get such types of problems, we should not use distributive property and expand the given equation. Avoid calculation mistakes based on sign conventions. Since the given equation has two factors, we will only get two values of v which will be real numbers.