# Solve the following quadratic equation by completing the square method ${{x}^{2}}+10x+21=0$.

Last updated date: 28th Mar 2023

•

Total views: 307.2k

•

Views today: 4.84k

Answer

Verified

307.2k+ 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

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE