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Last updated date: 29th Nov 2023
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MVSAT Dec 2023

Suppose that $f\left( x+3 \right)=3{{x}^{2}}+7x+4$ and $f\left( x \right)=a{{x}^{2}}+bx+c$ . What is ${{\left( a+b+c \right)}^{4}}$ ?

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Hint: Firstly, we have to consider the function $f\left( x+3 \right)=3{{x}^{2}}+7x+4$ . In this function, we have to replace x with $x+3$ . Then, we have to simplify the equation using algebraic identities and properties. We have to simplify this function and write it in the standard form of quadratic equations. We have to equate the resultant function and $f\left( x+3 \right)=3{{x}^{2}}+7x+4$ . Then, we have to compare the terms in ${{x}^{2}}$ , x and constant, that is, we have to equate the coefficients of ${{x}^{2}}$ , x and constant. From this step, we will get the value of a, b and c. Then, we have to substitute these values in the expression ${{\left( a+b+c \right)}^{4}}$ and simplify.

Complete step by step solution:
We are given that $f\left( x \right)=a{{x}^{2}}+bx+c$ . Let us replace x with $x+3$ in this function.
$\Rightarrow f\left( x+3 \right)=a{{\left( x+3 \right)}^{2}}+b\left( x+3 \right)+c$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ . Therefore, we can write the above equation as
  & \Rightarrow f\left( x+3 \right)=a\left( {{x}^{2}}+2\times 3\times x+{{3}^{2}} \right)+b\left( x+3 \right)+c \\
 & \Rightarrow f\left( x+3 \right)=a\left( {{x}^{2}}+6x+9 \right)+b\left( x+3 \right)+c \\
Let us apply distributive property in the above equation.
$\Rightarrow f\left( x+3 \right)=a{{x}^{2}}+6ax+9a+bx+3b+c$
We have to combine the like terms.
$\Rightarrow f\left( x+3 \right)=a{{x}^{2}}+\left( 6ax+bx \right)+\left( 9a+3b+c \right)$
Let us take the common term x from the second bracket outside.
$\Rightarrow f\left( x+3 \right)=a{{x}^{2}}+x\left( 6a+b \right)+\left( 9a+3b+c \right)...\left( i \right)$
We are given that $f\left( x+3 \right)=3{{x}^{2}}+7x+4$ . Let us equate this equation and the equation (i).
 $\Rightarrow 3{{x}^{2}}+7x+4=a{{x}^{2}}+x\left( 6a+b \right)+\left( 9a+3b+c \right)$
Let us compare the LHS and RHS. We have to compare the terms in ${{x}^{2}}$ , x and constant. Let us compare the term in ${{x}^{2}}$ . Their coefficients will be equal.
$\Rightarrow a=3...\left( ii \right)$
Now, we have to compare the terms in x. Their coefficients will be equal.
$\Rightarrow 6a+b=7$
We have to substitute the value of a from (ii) in the above equation.
  & \Rightarrow 6\times 3+b=7 \\
 & \Rightarrow 18+b=7 \\
Let us take 18 to the RHS.
  & \Rightarrow b=7-18 \\
 & \Rightarrow b=-11...\left( iii \right) \\
Now, we have to compare the constants.
$\Rightarrow 9a+3b+c=4$
Let us substitute the value of a and b from equations (i) and (ii) in the above equation.
  & \Rightarrow 9\times 3+3\times \left( -11 \right)+c=4 \\
 & \Rightarrow 27-33+c=4 \\
 & \Rightarrow -6+c=4 \\
Let us take -6 to the RHS.
  & \Rightarrow c=4+6 \\
 & \Rightarrow c=10...\left( iv \right) \\
Now, we have to find the value of ${{\left( a+b+c \right)}^{4}}$ . Let us substitute (ii), (iii) and (iv) in this expression.
  & \Rightarrow {{\left( a+b+c \right)}^{4}}={{\left( 3-11+10 \right)}^{4}} \\
 & \Rightarrow {{\left( a+b+c \right)}^{4}}={{2}^{4}} \\
 & \Rightarrow {{\left( a+b+c \right)}^{4}}=16 \\
Therefore, the value of ${{\left( a+b+c \right)}^{4}}$ is 16.

Note: Students must be thorough with algebraic properties and rules and must be able to simplify algebraic equations. We have obtained the equation (i) by rearranging the terms such that the function is in the form of a standard quadratic polynomial so that it is easy to compare with the given function $f\left( x+3 \right)$ .