
How do you find the value of c that makes $x^ {2} -13x+c$ into a perfect square?
Answer
519.3k+ views
Hint: To find out the value of c, that makes the given equation into a perfect square, we need to know what is a perfect square? Perfect square is a number, it can be expressed as the square of the number from a same number system. We can solve this question with the help of the ‘completing the square’ method.
Complete step by step solution:
In the given quadratic equation, the coefficients should be considered. So, now coefficient of x should be halved which would be \[-\dfrac{13}{2}\] and then it should be squared which forms to \[\dfrac{169}{4}\]which ‘c’ should be equal to \[{{\left( x-\dfrac{13}{2} \right)}^{2}}\].
In the given question \[{{x}^{2}}-13x+c\]\[..................\left( i \right)\]
we have the coefficients in the given quadratic equation as
\[\begin{align}
& a=1; \\
& b=-13; \\
& c=? \\
\end{align}\]
To find out the value of ‘c’, we need to follow the method called ‘completing the square’.
The given question is in the form of \[a{{x}^{2}}+bx+c\] (it is also called quadratic trinomial)
Quadratic trinomial is defined as “an expression in the form of \[a{{x}^{2}}+bx+c\], where x is a variable and a, b and c are non-real constants.
In the given problem, to make a perfect square, we should add on \[{{\left( \dfrac{b}{2} \right)}^{2}}\]as the c term, because we are following the completing the square method.
So, from the given equation, we have \[b=-13\]
Now substitute the value of b in equation (i)
We get, \[{{x}^{2}}-13x+{{\left( \dfrac{-13}{2} \right)}^{2}}\]is a perfect square \[{{x}^{2}}-13x+\left( 42.25 \right)\]
\[= \]\[{{\left( x-6.5 \right)}^{2}}\]
Then the value of c is \[{{\left( \dfrac{-13}{2} \right)}^{2}}= \left( 42.25 \right)\]
The value of c i.e., \[\left( 42.25 \right)\] makes the \[{{x}^{2}}-13x+\left( 42.25 \right)\] into a perfect square.
Note: While solving this question, the common mistake done by the students is they forget to divide the coefficient of the squared form of x, which leads to making everything incorrect. As the question is about perfect squares, students don’t understand how to write the trinomial as the form of a binomial square and also, they don’t understand how to use the property of square root.
Complete step by step solution:
In the given quadratic equation, the coefficients should be considered. So, now coefficient of x should be halved which would be \[-\dfrac{13}{2}\] and then it should be squared which forms to \[\dfrac{169}{4}\]which ‘c’ should be equal to \[{{\left( x-\dfrac{13}{2} \right)}^{2}}\].
In the given question \[{{x}^{2}}-13x+c\]\[..................\left( i \right)\]
we have the coefficients in the given quadratic equation as
\[\begin{align}
& a=1; \\
& b=-13; \\
& c=? \\
\end{align}\]
To find out the value of ‘c’, we need to follow the method called ‘completing the square’.
The given question is in the form of \[a{{x}^{2}}+bx+c\] (it is also called quadratic trinomial)
Quadratic trinomial is defined as “an expression in the form of \[a{{x}^{2}}+bx+c\], where x is a variable and a, b and c are non-real constants.
In the given problem, to make a perfect square, we should add on \[{{\left( \dfrac{b}{2} \right)}^{2}}\]as the c term, because we are following the completing the square method.
So, from the given equation, we have \[b=-13\]
Now substitute the value of b in equation (i)
We get, \[{{x}^{2}}-13x+{{\left( \dfrac{-13}{2} \right)}^{2}}\]is a perfect square \[{{x}^{2}}-13x+\left( 42.25 \right)\]
\[= \]\[{{\left( x-6.5 \right)}^{2}}\]
Then the value of c is \[{{\left( \dfrac{-13}{2} \right)}^{2}}= \left( 42.25 \right)\]
The value of c i.e., \[\left( 42.25 \right)\] makes the \[{{x}^{2}}-13x+\left( 42.25 \right)\] into a perfect square.
Note: While solving this question, the common mistake done by the students is they forget to divide the coefficient of the squared form of x, which leads to making everything incorrect. As the question is about perfect squares, students don’t understand how to write the trinomial as the form of a binomial square and also, they don’t understand how to use the property of square root.
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