Question

How do you solve \[25{{x}^{2}}-30x=-9\]?

Answer 1

Hint: In order to solve the given question \[25{{x}^{2}}-30x=-9\] and find the value of variable \[x\], where a “variable” is a symbol which functions as a placeholder for varying expression or quantities. First move all the terms from right-hand-side to left side. After this simplify the quadratic equation with the help of splitting the middle term then take the common factors from the two pairs in the brackets. Now, at last rearrange and isolate the variable \[x\] to find each solution.

Complete step by step solution:
According to the question, given equation is as follows:
\[25{{x}^{2}}-30x=-9\]
To solve the above equation, first move all the terms from right-hand side to the left-hand side, we get:
\[\Rightarrow 25{{x}^{2}}-30x-\left( -9 \right)=0\]
Solve the bracket, by applying the concept of number system that is two 'minuses' make a plus in the above equation, we get:
\[\Rightarrow 25{{x}^{2}}-30x+9=0\]
As you can see it’s a quadratic equation, and to solve quadratic equation we apply split of the middle term rule or in other words sum-product pattern in the above equation, we get:
\[\Rightarrow 25{{x}^{2}}-15x-15x+9=0\]
After this take the common factors from the two pairs in the brackets, we get:
\[\Rightarrow 5x\left( 5x-3 \right)-3\left( 5x-3 \right)=0\]
Now rewrite the expression in the left-hand side in the factored form, we get:
\[\Rightarrow \left( 5x-3 \right)\left( 5x-3 \right)=0\]
As you can see both the terms in brackets are similar therefore, we can add their powers that is we can write \[\left( 5x-3 \right)\left( 5x-3 \right)={{\left( 5x-3 \right)}^{1+1}}={{\left( 5x-3 \right)}^{2}}\]. Now applying this in the above equation we get:
\[\Rightarrow {{\left( 5x-3 \right)}^{2}}=0\]
Simplify it further by creating separate equation from above equation, rearrange it and isolate the variable \[x\] to find the solution, we get:
\[\Rightarrow \left( 5x-3 \right)=0\] and \[\left( 5x-3 \right)=0\]
\[\Rightarrow 5x=3\] and \[5x=3\]
\[\Rightarrow x=\dfrac{3}{5},\dfrac{3}{5}\]
Therefore, the values for variable \[x\] after solving the given equation \[25{{x}^{2}}-30x=-9\] are \[\dfrac{3}{5}\]and \[\dfrac{3}{5}\].

Note: Students generally make mistakes while applying the concept of splitting the middle term to solve the quadratic term, they make mistakes in identifying the common terms and take wrong terms in common which further leads to the wrong answer. Key point is to remember this concept of splitting the middle term that is in quadratic factorization, Splitting of Middle Term is variable term is the sum of two factors and product equal to last term and the concept of number system that is two 'minuses' make a plus.