Answer

Verified

406.8k+ views

Hint: We are required to use completing the square method to solve this question. The first step is to make the coefficient of ${{x}^{2}}$ equal to 1 by dividing its coefficient throughout the equation. Then, we can transpose the constant term to the right- hand side.

Complete step-by-step answer:

Here, we have the given equation as ${{x}^{2}}+10x+21=0$.

Since we have to use completing the square method, there are steps to be followed. The step by step process can be defined in general for an equation $a{{x}^{2}}+bx+c=0$ as,

1. Firstly, we have to divide all the terms by the coefficient of ${{x}^{2}}$, i.e., $a$.

2. Then, we have to transpose the term $\frac{c}{a}$ to the right-hand side of the equation.

3. Add the required squared term on both the sides, to get the whole square form to the left side of the equation.

4. Thus, we have converted the left-side terms in a whole square form and have to take square root on both sides.

5. We get two values on the right side, positive and negative values of the right-side equation.

6. Equating the right-side term with $x$, we get the values of $x$.

Now, applying the same steps on the given equation, we get

$\begin{align}

& \Rightarrow {{x}^{2}}+10x+21=0 \\

& \Rightarrow \frac{{{x}^{2}}+10x+21}{1}=\frac{0}{1} \\

& \Rightarrow {{x}^{2}}+10x+21=0 \\

& \Rightarrow {{x}^{2}}+10x=-21 \\

& \Rightarrow {{x}^{2}}+2\left( 5 \right)\left( x \right)=-21 \\

& \Rightarrow {{x}^{2}}+2\left( 5 \right)\left( x \right)+{{5}^{2}}=-21+{{5}^{2}} \\

& \Rightarrow {{x}^{2}}+2\left( 5 \right)\left( x \right)+{{5}^{2}}=-21+25 \\

& \Rightarrow {{x}^{2}}+2\left( 5 \right)\left( x \right)+{{5}^{2}}=4 \\

\end{align}$

Now, the left-side of the above equation is in the form of ${{a}^{2}}+2ab+{{b}^{2}}$, which is equal to ${{\left( a+b \right)}^{2}}$, i.e.,

$\begin{align}

& \Rightarrow {{x}^{2}}+2\left( 5 \right)\left( x \right)+{{5}^{2}}=4 \\

& \Rightarrow {{\left( x+5 \right)}^{2}}=4 \\

\end{align}$

Now, taking square root on both sides, we get

$\begin{align}

& \Rightarrow {{\left( x+5 \right)}^{2}}=4 \\

& \Rightarrow \sqrt{{{\left( x+5 \right)}^{2}}}=\sqrt{4} \\

& \Rightarrow \left( x+5 \right)=\pm 2 \\

\end{align}$

Considering the positive value on right-side, we get

$\begin{align}

& \Rightarrow x+5=2 \\

& \Rightarrow x=2-5 \\

& \Rightarrow x=-3 \\

\end{align}$

Considering the negative value on right-side, we get

$\begin{align}

& \Rightarrow x+5=-2 \\

& \Rightarrow x=-2-5 \\

& \Rightarrow x=-7 \\

\end{align}$

Hence, on solving the given equation by completing the square method, we get values of $x=-2$ and $x=-7$.

Note: After obtaining the answer, the student can substitute it in the given equation to check if it satisfies the given equation. The student can also cross-check the obtained answer by solving the given equation using the middle term split method or the quadratic formula method, but that will lead to a waste of time.

Complete step-by-step answer:

Here, we have the given equation as ${{x}^{2}}+10x+21=0$.

Since we have to use completing the square method, there are steps to be followed. The step by step process can be defined in general for an equation $a{{x}^{2}}+bx+c=0$ as,

1. Firstly, we have to divide all the terms by the coefficient of ${{x}^{2}}$, i.e., $a$.

2. Then, we have to transpose the term $\frac{c}{a}$ to the right-hand side of the equation.

3. Add the required squared term on both the sides, to get the whole square form to the left side of the equation.

4. Thus, we have converted the left-side terms in a whole square form and have to take square root on both sides.

5. We get two values on the right side, positive and negative values of the right-side equation.

6. Equating the right-side term with $x$, we get the values of $x$.

Now, applying the same steps on the given equation, we get

$\begin{align}

& \Rightarrow {{x}^{2}}+10x+21=0 \\

& \Rightarrow \frac{{{x}^{2}}+10x+21}{1}=\frac{0}{1} \\

& \Rightarrow {{x}^{2}}+10x+21=0 \\

& \Rightarrow {{x}^{2}}+10x=-21 \\

& \Rightarrow {{x}^{2}}+2\left( 5 \right)\left( x \right)=-21 \\

& \Rightarrow {{x}^{2}}+2\left( 5 \right)\left( x \right)+{{5}^{2}}=-21+{{5}^{2}} \\

& \Rightarrow {{x}^{2}}+2\left( 5 \right)\left( x \right)+{{5}^{2}}=-21+25 \\

& \Rightarrow {{x}^{2}}+2\left( 5 \right)\left( x \right)+{{5}^{2}}=4 \\

\end{align}$

Now, the left-side of the above equation is in the form of ${{a}^{2}}+2ab+{{b}^{2}}$, which is equal to ${{\left( a+b \right)}^{2}}$, i.e.,

$\begin{align}

& \Rightarrow {{x}^{2}}+2\left( 5 \right)\left( x \right)+{{5}^{2}}=4 \\

& \Rightarrow {{\left( x+5 \right)}^{2}}=4 \\

\end{align}$

Now, taking square root on both sides, we get

$\begin{align}

& \Rightarrow {{\left( x+5 \right)}^{2}}=4 \\

& \Rightarrow \sqrt{{{\left( x+5 \right)}^{2}}}=\sqrt{4} \\

& \Rightarrow \left( x+5 \right)=\pm 2 \\

\end{align}$

Considering the positive value on right-side, we get

$\begin{align}

& \Rightarrow x+5=2 \\

& \Rightarrow x=2-5 \\

& \Rightarrow x=-3 \\

\end{align}$

Considering the negative value on right-side, we get

$\begin{align}

& \Rightarrow x+5=-2 \\

& \Rightarrow x=-2-5 \\

& \Rightarrow x=-7 \\

\end{align}$

Hence, on solving the given equation by completing the square method, we get values of $x=-2$ and $x=-7$.

Note: After obtaining the answer, the student can substitute it in the given equation to check if it satisfies the given equation. The student can also cross-check the obtained answer by solving the given equation using the middle term split method or the quadratic formula method, but that will lead to a waste of time.

Recently Updated Pages

Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

Give simple chemical tests to distinguish between the class 12 chemistry CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

How will you bring about the following conversions class 12 chemistry CBSE

Consider a system of two identical particles One of class 11 physics CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

State the laws of reflection of light