Let the function \[f\left( x \right) = {x^2} + x - 6\], For what values of \[t\] is \[f\left( {t - 5} \right) = 0\]?
(A) \[ - 3\] and \[2\]
(B) \[ - 2\] and \[3\]
(C) \[5\]
(D) \[2\] and \[7\]
Answer
649.8k+ views
Hint: In this question, we have to evaluate the value of \[t\] when the functional value is equal to zero.
We need to first put \[\left( {t - 5} \right)\] in the given function as we need to find out the value of \[t\] when the functional value is equal to zero, then we equal it to zero. After that, we will solve the quadratic equation to get the values of \[t\].
Formula used:
Here we have used the algebraic formula,
\[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\].
Complete step by step answer:
It is given that, the function \[f(x)\] is defined as \[f\left( x \right) = {x^2} + x - 6\].
We need to find out the values of \[t\] for which \[f\left( {t - 5} \right) = 0\].
Now we will put \[\left( {t - 5} \right)\] in the function \[f\left( x \right) = {x^2} + x - 6\] we get,
\[ \Rightarrow f\left( {t - 5} \right) = {\left( {t - 5} \right)^2} + \left( {t - 5} \right) - 6\]
Using the algebraic formula, \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\], We get,
\[ \Rightarrow {t^2} - 2 \times t \times 5 + 25 + t - 5 - 6\]
We will get by arranging term by term in this quadratic equation,
\[ \Rightarrow {t^2} - 10t + t + 14\]
Subtracting the terms we get,
\[ \Rightarrow {t^2} - 9t + 14\]
Solving the quadratic equation by middle term factor we get,
\[ \Rightarrow {t^2} - 7t - 2t + 14\]
Taking common \[t\] from first two terms and \[ - 2\] last two terms,
\[ \Rightarrow t\left( {t - 7} \right) - 2\left( {t - 7} \right)\]
Factorizing we get,
\[ \Rightarrow \left( {t - 7} \right)\left( {t - 2} \right)\]
Also it is given that, \[f\left( {t - 5} \right) = 0\].
Thus we get, \[\left( {t - 7} \right)\left( {t - 2} \right) = 0\].
Equating two factors equal to zero,
\[ \Rightarrow \left( {t - 7} \right) = 0\] and \[\left( {t - 2} \right) = 0\]
\[ \Rightarrow t = 7\] and \[t = 2\]
For the value of \[t\] is \[2\] and \[7\], \[f\left( {t - 5} \right) = 0\]
$\therefore $ Option (D) is the correct option.
Note:
Like this problem, we have concentrated on the step of factorization. In mathematics there are so many methods to factorize an equation. Here we use the middle term process. So we have to concentrate on splitting the middle term either addition of two numbers or subtraction of two numbers, then make the quadratic equation as the multiplication of two factors. We may make mistakes on that splitting.
The quadratic equation can be solved by the middle-term process. Splitting the middle term either addition of two numbers or subtraction of two numbers, then make the quadratic equation as the multiplication of two factors. By solving the two factors we will get the solution.
We need to first put \[\left( {t - 5} \right)\] in the given function as we need to find out the value of \[t\] when the functional value is equal to zero, then we equal it to zero. After that, we will solve the quadratic equation to get the values of \[t\].
Formula used:
Here we have used the algebraic formula,
\[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\].
Complete step by step answer:
It is given that, the function \[f(x)\] is defined as \[f\left( x \right) = {x^2} + x - 6\].
We need to find out the values of \[t\] for which \[f\left( {t - 5} \right) = 0\].
Now we will put \[\left( {t - 5} \right)\] in the function \[f\left( x \right) = {x^2} + x - 6\] we get,
\[ \Rightarrow f\left( {t - 5} \right) = {\left( {t - 5} \right)^2} + \left( {t - 5} \right) - 6\]
Using the algebraic formula, \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\], We get,
\[ \Rightarrow {t^2} - 2 \times t \times 5 + 25 + t - 5 - 6\]
We will get by arranging term by term in this quadratic equation,
\[ \Rightarrow {t^2} - 10t + t + 14\]
Subtracting the terms we get,
\[ \Rightarrow {t^2} - 9t + 14\]
Solving the quadratic equation by middle term factor we get,
\[ \Rightarrow {t^2} - 7t - 2t + 14\]
Taking common \[t\] from first two terms and \[ - 2\] last two terms,
\[ \Rightarrow t\left( {t - 7} \right) - 2\left( {t - 7} \right)\]
Factorizing we get,
\[ \Rightarrow \left( {t - 7} \right)\left( {t - 2} \right)\]
Also it is given that, \[f\left( {t - 5} \right) = 0\].
Thus we get, \[\left( {t - 7} \right)\left( {t - 2} \right) = 0\].
Equating two factors equal to zero,
\[ \Rightarrow \left( {t - 7} \right) = 0\] and \[\left( {t - 2} \right) = 0\]
\[ \Rightarrow t = 7\] and \[t = 2\]
For the value of \[t\] is \[2\] and \[7\], \[f\left( {t - 5} \right) = 0\]
$\therefore $ Option (D) is the correct option.
Note:
Like this problem, we have concentrated on the step of factorization. In mathematics there are so many methods to factorize an equation. Here we use the middle term process. So we have to concentrate on splitting the middle term either addition of two numbers or subtraction of two numbers, then make the quadratic equation as the multiplication of two factors. We may make mistakes on that splitting.
The quadratic equation can be solved by the middle-term process. Splitting the middle term either addition of two numbers or subtraction of two numbers, then make the quadratic equation as the multiplication of two factors. By solving the two factors we will get the solution.
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