Question

How do you solve \[7{{r}^{2}}-14r=-7\] by factoring?

Answer 1

Hint: We are given \[7{{r}^{2}}-14r=-7\] to solve this we learn about quadratic equation, the number of solutions of the quadratic equation. We will learn how to factor the quadratic equation, we will simplify by taking the common terms out then we use the zero product rule to get our answer. At the last, we will also learn about quadratic formulas for solving such equations in an easy way and more speedy way.

Complete step by step answer:
We are given in the question \[7{{r}^{2}}-14r=-7\] and we are asked to solve the given problem. First, we observe that it has a maximum power of ‘2’ so it is a quadratic equation. Now we should know that the quadratic equation has 2 solutions or we say an equation of power ‘n’ will have an ‘n’ solution. Now as it is a quadratic equation we will change it into standard form \[a{{x}^{2}}+bx+c=0.\]
So, we are given
\[7{{r}^{2}}-14r=-7\]
Adding 7 to both sides we get
\[\Rightarrow 7{{r}^{2}}-14r+7=0-7+7\]
\[\Rightarrow 7{{r}^{2}}-14r+7=0\]
Now, we have to solve the equation \[7{{r}^{2}}-14r+7=0.\]
To solve this equation we first take the greatest common factor possibly available to the terms. As we can see that in \[7{{r}^{2}}-14r+7=0,\] 7 is common. So, we take it out from all the terms, we get,
\[\Rightarrow 7{{r}^{2}}-14r+7=7\left( {{r}^{2}}-2r+1 \right)\]
To solve the equation, we will use the middle term split. The middle term split method say for any quadratic equation is
1. We first product \[a\times c\]
2. We try to find the number such that their sum or difference is ‘b’ while the product is the same as \[a\times c.\]
3. We use that and separate like terms.
4. We use the zero product rule to get our solution.
Now we have \[{{r}^{2}}-2r+1.\] So, we have a = 1, b = – 2 and c = 1. So, \[a\times c=1.\]
Now, we can see that \[1\times 1=1.\] And also 1 + 1 = 2. So, we use this to split the middle term ‘b’. So, \[{{r}^{2}}-2r+1\] becomes
\[{{r}^{2}}-\left( 1+1 \right)r+1\]
By simplifying, we get,
\[\Rightarrow {{r}^{2}}-r-r+1\]
Now, take the common from the first 2 terms and the last 2 terms, we get,
\[\Rightarrow r\left( r-1 \right)-1\left( r-1 \right)\]
Simplifying further, we get,
\[\Rightarrow \left( r-1 \right)\left( r-1 \right)\]
Now using zero product rules, we have,
\[\Rightarrow r-1=0;r-1=0\]
So, we get,
\[\Rightarrow r=1;r=1\]
So, we get r = 1 as the solution.

Note:
Another way to solve this is we will use the quadratic formula. The quadratic formula states that for any \[a{{x}^{2}}+bx+c=0,\] the solution is given as \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.\] For \[7\left( {{r}^{2}}-2r+1 \right),\] we have a = 1, b = – 2 and c = 1. So, using these, we get,
\[r=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\times 1\times 1}}{2\times 1}\]
\[\Rightarrow r=\dfrac{-\left( -2 \right)\pm \sqrt{4-4}}{2}\]
So, simplifying we get
\[\Rightarrow r=\dfrac{2\pm 0}{2}\]
\[\Rightarrow r=1\]
So, we have the same solution both the times, so r = 1 in the solution of \[7\left( {{r}^{2}}-2r+1 \right),\] so we get, r = 1 in the solution of \[7{{r}^{2}}-14r=-7.\]