Q12 PT. - Algebra 1

Question

Solve the given equation and check the solution.

$\sqrt{4x-3}=6-x$

Step-by-Step Solution

Answer

The solution of $\sqrt{4x-3}=6-x$ is 3.

Step 1. State the concept of ‘square of a square root’.

When the square root of a term is squared, then the resulting value will be the term which is inside the root. That is,

${\left(\sqrt{a}\right)}^2=a$

Step 2. State the concept of ‘quadratic equation’.

The equation which consists of only one variable and whose highest degree is ‘2’, is  called as quadratic equation.

Step 3. Define ‘Factorization method’.

The method of writing polynomials into its simplest form, that is writing it as the product of factors is called as factorization method.

Step 4. Find the value of ‘ x ’.

Solve the equation $\sqrt{4x-3}=6-x$ to get the value of ‘x’.

$\sqrt{4x-3}=6-x$ 

 

Squaring both the sides

$$\begin{gathered} {\left(\sqrt{4x-3}\right)}^2={\left(6-x\right)}^2 \\ 4x-3={\left(6\right)}^2-2\left(6\right)\left(x\right)+{\left(x\right)}^2\left[\text{Using }{\left(\sqrt{a}\right)}^2=a\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}{(a-b)}^2=a^2-2ab+b^2\right] \\ 4x-3=36-12x+x^2 \\ 4x-3=x^2-12x+36 \\ x^2-12x+36=4x-3 \\ {x}^2-12x+36-4x+3=0 \\ {x}^2-12x-4x+36+3=0 \\ x^2-16x+39=0 \\ x^2-13x-3x+39=0\left[\text{Using\hspace{0.17em}\hspace{0.17em}factorization\hspace{0.17em}\hspace{0.17em}method}\right] \\ x\left(x-13\right)-3\left(x-13\right)=0 \\ \left(x-13\right)\left(x-3\right)=0 \end{gathered}$$ 

$$\begin{gathered} \left(x-13\right)=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(x-3\right)=0\left[\text{If\hspace{0.17em}\hspace{0.17em}}ab=0,\text{\hspace{0.17em}\hspace{0.17em}then\hspace{0.17em}\hspace{0.17em}either\hspace{0.17em}\hspace{0.17em}}a=0\text{\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}}b=0\right] \\ x-13=0\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x-3=0 \\ x=13\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x=3 \end{gathered}$$ 

Step 5. Verify the solution.

Consider $x=13$

 

Substitute $x=13$ in the left hand side(L.H.S) of the equation $\sqrt{4x-3}=6-x$. 

$$\begin{gathered} \text{L.H.S}=\sqrt{4x-3} \\ =\sqrt{4\left(13\right)-3} \\ =\sqrt{52-3} \\ =\sqrt{49} \\ =7\dots \left(1\right) \end{gathered}$$ 

 

Now, Substitute $x=13$ in the right hand side(R.H.S) of the equation $\sqrt{4x-3}=6-x$. 

$$\begin{gathered} \text{R.H.S}=6-x \\ =6-13 \\ =-7\dots \left(2\right) \end{gathered}$$ 

From (1) and (2)

$\text{L.H.S}\ne \text{R.H.S}$

Thus, $x=13$ does not satisfies the given equation.

 

Consider $x=13$

 

Substitute $x=13$ in the left hand side(L.H.S) of the equation $\sqrt{4x-3}=6-x$. 

$$\begin{gathered} \text{L.H.S}=\sqrt{4x-3} \\ =\sqrt{4\left(3\right)-3} \\ =\sqrt{12-3} \\ =\sqrt{9} \\ =3\dots \left(3\right) \end{gathered}$$ 

Now, Substitute $x=3$ in the right hand side(R.H.S) of the equation $\sqrt{4x-3}=6-x$. 

$$\begin{gathered} \text{R.H.S}=6-x \\ =6-3 \\ =3\dots \left(4\right) \end{gathered}$$ 

From (3) and (4)

$\text{L.H.S=R.H.S}$

$x=3$ satisfies the given equation.

Therefore, $x=3$ is the solution of the given equation.