# Solve for $a,b$ : ${{a}^{2}}+{{b}^{2}}-4a+16b+68=0$

Last updated date: 26th Mar 2023

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Answer

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Hint: The given question is related to algebraic identities. Express the given expression in the form of ${{\left( p+q \right)}^{2}}$ and ${{\left( x-y \right)}^{2}}$, and use the fact that “ if the sum of squares of two numbers is equal zero, then both numbers must be equal to zero ”, to find the values of $a$ and $b$.

Complete step-by-step answer:

Before proceeding with the solution, let’s write down the algebraic identities used to solve the given problem.

${{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}$

${{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}$

We know that the expression ${{x}^{2}}+2xy+{{y}^{2}}$ can be written as ${{\left( x+y \right)}^{2}}$. Also, we can write ${{x}^{2}}-2xy+{{y}^{2}}$ as ${{\left( x-y \right)}^{2}}$ . And, we can write $68$ as $64+4$ .

Now, we are given the expression ${{a}^{2}}+{{b}^{2}}-4a+16b+68=0$. We know we can write $68$ as $64+4$ . So, we can rewrite the given expression as ${{a}^{2}}+{{b}^{2}}-4a+16b+64+4=0$. Now, we will rearrange the terms. On rearranging the terms, we get \[~{{a}^{2}}-4a+4+{{b}^{2}}+16b+64=0\] . Now, we know, $4$ can be rewritten as ${{2}^{2}}$ , $4a$ can be rewritten as $2\times a\times 2$ , $64$ can be rewritten as ${{8}^{2}}$ and $16b$ can be rewritten as $2\times b\times 8$ . So, we can rewrite the given expression as ${{a}^{2}}-\left( 2\times a\times 2 \right)+{{2}^{2}}+{{b}^{2}}+\left( 2\times b\times 8 \right)+{{8}^{2}}=0$ . Now, the first three terms of the expression are of the form ${{x}^{2}}-2xy+{{y}^{2}}$ and the next three terms of the expression are of the form ${{x}^{2}}+2xy+{{y}^{2}}$ . But we know, ${{x}^{2}}-2xy+{{y}^{2}}$ can be written as ${{\left( x+y \right)}^{2}}$ and ${{x}^{2}}-2xy+{{y}^{2}}$ can be written as ${{\left( x-y \right)}^{2}}$ . So, we can rewrite the expression as ${{\left( a-2 \right)}^{2}}+{{\left( b+8 \right)}^{2}}=0$ . We can see that both the terms of the expression are perfect squares. We know that a square of a number can never be negative. So, if the sum of squares of two numbers is zero, then both the numbers must be zero. So, $a-2=0$ and $b+8=0$. Now, $a-2=0$ .

$\Rightarrow a=2$

And, $b+8=0$.

$\Rightarrow b=-8$

So, the values of $a$ and $b$ for which ${{a}^{2}}+{{b}^{2}}-4a+16b+68=0$ are $a=2$ and $b=-8$ .

Note: While rearranging the terms, take care of the sign of the terms. While rearranging, sign mistakes are common.

Complete step-by-step answer:

Before proceeding with the solution, let’s write down the algebraic identities used to solve the given problem.

${{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}$

${{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}$

We know that the expression ${{x}^{2}}+2xy+{{y}^{2}}$ can be written as ${{\left( x+y \right)}^{2}}$. Also, we can write ${{x}^{2}}-2xy+{{y}^{2}}$ as ${{\left( x-y \right)}^{2}}$ . And, we can write $68$ as $64+4$ .

Now, we are given the expression ${{a}^{2}}+{{b}^{2}}-4a+16b+68=0$. We know we can write $68$ as $64+4$ . So, we can rewrite the given expression as ${{a}^{2}}+{{b}^{2}}-4a+16b+64+4=0$. Now, we will rearrange the terms. On rearranging the terms, we get \[~{{a}^{2}}-4a+4+{{b}^{2}}+16b+64=0\] . Now, we know, $4$ can be rewritten as ${{2}^{2}}$ , $4a$ can be rewritten as $2\times a\times 2$ , $64$ can be rewritten as ${{8}^{2}}$ and $16b$ can be rewritten as $2\times b\times 8$ . So, we can rewrite the given expression as ${{a}^{2}}-\left( 2\times a\times 2 \right)+{{2}^{2}}+{{b}^{2}}+\left( 2\times b\times 8 \right)+{{8}^{2}}=0$ . Now, the first three terms of the expression are of the form ${{x}^{2}}-2xy+{{y}^{2}}$ and the next three terms of the expression are of the form ${{x}^{2}}+2xy+{{y}^{2}}$ . But we know, ${{x}^{2}}-2xy+{{y}^{2}}$ can be written as ${{\left( x+y \right)}^{2}}$ and ${{x}^{2}}-2xy+{{y}^{2}}$ can be written as ${{\left( x-y \right)}^{2}}$ . So, we can rewrite the expression as ${{\left( a-2 \right)}^{2}}+{{\left( b+8 \right)}^{2}}=0$ . We can see that both the terms of the expression are perfect squares. We know that a square of a number can never be negative. So, if the sum of squares of two numbers is zero, then both the numbers must be zero. So, $a-2=0$ and $b+8=0$. Now, $a-2=0$ .

$\Rightarrow a=2$

And, $b+8=0$.

$\Rightarrow b=-8$

So, the values of $a$ and $b$ for which ${{a}^{2}}+{{b}^{2}}-4a+16b+68=0$ are $a=2$ and $b=-8$ .

Note: While rearranging the terms, take care of the sign of the terms. While rearranging, sign mistakes are common.

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