How do you solve \[x - 12 = \sqrt {16x} \] ?
Answer
596.4k+ views
Hint: In this question, we have to solve the given equation. First we need to remove the square root sign of the right hand side by squaring both sides. Then rearranging the terms we will get a quadratic equation. Solving this equation by middle term method we will get the required solution.
Formula used: \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
Complete step-by-step solution:
It is given that, \[x - 12 = \sqrt {16x} \ldots \ldots .\left( 1 \right)\]
We need to solve the given equation.
For solving the equation first we need to remove the square root sign of the right hand side by squaring both sides.
Squaring both sides of \[\left( 1 \right)\] we get,
\[ \Rightarrow {\left( {x - 12} \right)^2} = 16x\]
Solving we get,
\[ \Rightarrow {x^2} - 24x + 144 = 16x\] [Using the formula \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\] ]
Rearranging we get,
\[ \Rightarrow {x^2} - 24x - 16x + 144 = 0\]
Let us add the term and we get
\[ \Rightarrow {x^2} - 40x + 144 = 0\]
Solving the quadratic equation by solving middle term method we get,
\[ \Rightarrow {x^2} - 36x - 4x + 144 = 0\]
Taking \[x\] as common on first two terms and \[ - 4\] as common on third and fourth term we get,
\[ \Rightarrow x\left( {x - 36} \right) - 4\left( {x - 36} \right) = 0\]
Taking common term we get,
\[ \Rightarrow \left( {x - 36} \right)\left( {x - 4} \right) = 0\]
Then \[x = 36\] and \[x = 4\]
Hence we get, the value of \[x\] is either \[36\] or \[4\]
Note: Quadratic equation: In algebra a quadratic equation is any equation that can be rearranged in standard form as \[a{x^2} + bx + c = 0\] , where x represents an unknown and a, b, c represent known numbers where \[a \ne 0\] .
If \[a = 0\] then it will become a linear equation not quadratic as there is no \[a{x^2}\] term.
Quadratic equations can be solved by a middle term process.
In this method we will Split the middle term in such a way that either it will be represented as addition of two numbers or subtraction of two numbers, then making the quadratic equation as a multiplication of two factors and solving the two factors we will get the solution.
Formula used: \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
Complete step-by-step solution:
It is given that, \[x - 12 = \sqrt {16x} \ldots \ldots .\left( 1 \right)\]
We need to solve the given equation.
For solving the equation first we need to remove the square root sign of the right hand side by squaring both sides.
Squaring both sides of \[\left( 1 \right)\] we get,
\[ \Rightarrow {\left( {x - 12} \right)^2} = 16x\]
Solving we get,
\[ \Rightarrow {x^2} - 24x + 144 = 16x\] [Using the formula \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\] ]
Rearranging we get,
\[ \Rightarrow {x^2} - 24x - 16x + 144 = 0\]
Let us add the term and we get
\[ \Rightarrow {x^2} - 40x + 144 = 0\]
Solving the quadratic equation by solving middle term method we get,
\[ \Rightarrow {x^2} - 36x - 4x + 144 = 0\]
Taking \[x\] as common on first two terms and \[ - 4\] as common on third and fourth term we get,
\[ \Rightarrow x\left( {x - 36} \right) - 4\left( {x - 36} \right) = 0\]
Taking common term we get,
\[ \Rightarrow \left( {x - 36} \right)\left( {x - 4} \right) = 0\]
Then \[x = 36\] and \[x = 4\]
Hence we get, the value of \[x\] is either \[36\] or \[4\]
Note: Quadratic equation: In algebra a quadratic equation is any equation that can be rearranged in standard form as \[a{x^2} + bx + c = 0\] , where x represents an unknown and a, b, c represent known numbers where \[a \ne 0\] .
If \[a = 0\] then it will become a linear equation not quadratic as there is no \[a{x^2}\] term.
Quadratic equations can be solved by a middle term process.
In this method we will Split the middle term in such a way that either it will be represented as addition of two numbers or subtraction of two numbers, then making the quadratic equation as a multiplication of two factors and solving the two factors we will get the solution.
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