Question

How do you evaluate \[3\left( {{x}^{2}}-1 \right)-\left( {{x}^{2}}-7x+10 \right)\]?

Answer 1

Hint:This is the question of algebraic expression as it consists of variables, coefficients, constants, and mathematical operations such as addition, subtraction, multiplication and division. In the given question of an expression, you just need to simplify the expression by using mathematical operations and evaluate further.

Formula used:
The quadratic formula provides the solution for the quadratic equation:
\[a{{x}^{2}}+bx+c=0\]
In which a, b and c are the coefficient of respectively terms in the quadratic equation, as follows:
Roots of the quadratic equation= \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]

Complete step by step answer:
We have the given expression:
\[3\left( {{x}^{2}}-1 \right)-\left( {{x}^{2}}-7x+10 \right)\]
First, we need to multiply 3 to \[{{x}^{2}}-1\], we get
\[3{{x}^{2}}-3-\left( {{x}^{2}}-7x+10 \right)\]
Open the bracket and distribute the negative sign to \[\left( {{x}^{2}}-7x+10 \right)\], we get
\[3{{x}^{2}}-3-{{x}^{2}}+7x-10\]
Combine the like terms in an expression, we get
\[2{{x}^{2}}+7x-13\]
Trying to factorize the expression by splitting the middle term, and hence we find it is not possible i.e. no two such factors can be found. Now, we solve the resultant quadratic equation by quadratic formula: Given quadratic equation,
\[2{{x}^{2}}+7x-13\]
The quadratic formula provides the solution for the quadratic equation:
\[a{{x}^{2}}+bx+c=0\]
In which a, b and c are the coefficient of respectively terms in the quadratic equation, as follows:
Roots of the quadratic equation= \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Determine the quadratic equation’s coefficients a, b and c:
The coefficient of the given quadratic equation \[2{{x}^{2}}+7x-13\] are,
$a = 2\\
\Rightarrow b = 7\\
\Rightarrow c = -13$

Plug these coefficient into the quadratic formula:
\[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-7\pm \sqrt{{{7}^{2}}-\left( 4\times 2\times \left( -13 \right) \right)}}{2\times 2}\]
Solve exponents and square root, we get
\[\dfrac{-7\pm \sqrt{{{7}^{2}}-\left( 4\times 2\times \left( -13 \right) \right)}}{2\times 2}\]
\[\Rightarrow\dfrac{-7\pm \sqrt{49-\left( 4\times 2\times \left( -13 \right) \right)}}{2\times 2}\]
Performing any multiplication and division given in the formula,
\[\dfrac{-7\pm \sqrt{49-\left( 8\times \left( -13 \right) \right)}}{4}\]
\[\Rightarrow\dfrac{-7\pm \sqrt{49-\left( -104 \right)}}{4}\]
\[\Rightarrow\dfrac{-7\pm \sqrt{49+104}}{4}\]
\[\Rightarrow\dfrac{-7\pm \sqrt{153}}{4}\]

To get the result,
Let \[x=\dfrac{-7\pm \sqrt{153}}{4}\],
Simplifying the \[\sqrt{153}\], we get
The prime factorization is \[{{3}^{2}}\times 17\]
Therefore,
\[\sqrt{153}=3\times \sqrt{17}\]
Now, solve the equation for \[x\],
\[x=\dfrac{-7\pm 3\times \sqrt{17}}{4}\]
Possible value of \[x\] are,
\[x=\dfrac{-7+3\times \sqrt{17}}{4}\] and \[x=\dfrac{-7-3\times \sqrt{17}}{4}\]
\[\Rightarrow x=\dfrac{-7+3\times 4.123}{4}\] and \[x=\dfrac{-7-3\times 4.123}{4}\]
\[\Rightarrow x=\dfrac{-7+12.369}{4}\] and \[x=\dfrac{-7-12.369}{4}\]
\[\Rightarrow x=\dfrac{5.369}{4}\] and \[x=\dfrac{-19.369}{4}\]
\[x=1.342\] and \[x=-4.842\]

Therefore, the possible values of \[x\] are 1.342 and -4.842.

Note:To solve or evaluation these types of expression, we need to know about the:
Solving quadratic equations using the formula
-Simplifying radicals
-Find prime factors
-The general form of quadratic equation is \[a{{x}^{2}}+bx+c=0\], where a, b and c are the numerical coefficients or constants, and the value of \[x\] is unknown one fundamental rule is that the value of a, the first constant can never be zero.