Answer

Verified

433.2k+ views

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.

Recently Updated Pages

The branch of science which deals with nature and natural class 10 physics CBSE

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 can I graph Charles law class 11 chemistry CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

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

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

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

Tin is extracted from its ore cassiterite by A Electrolytic class 12 chemistry CBSE

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