How do you solve \[7{{x}^{2}}-6=57\]?
Answer
620.4k+ views
Hint: We will solve this expression by first writing the expression in terms of x, and for this we will use arithmetic operations and then we will solve it for x. The polynomial expression given to us is a quadratic equation, so we will have two values of x satisfying the given expression.
Complete step by step solution:
According to the given question, we have to solve in terms of x. The given polynomial expression is of the degree 3 and so it is a quadratic equation. The given equation will have two values which when substituted in the equation will satisfy the condition.
We will start first with rearranging the equation, we will separate the x components and the constant terms and then proceed with solving for x.
We have,
\[7{{x}^{2}}-6=57\]
On the left hand side (LHS) of the equality, we have a negative number 6, so we in order to separate the x components and the constant terms, we will add 6 on both the sides of the equality, we get
\[\Rightarrow (7{{x}^{2}}-6)+6=57+6\]
So, the 6 will get cancelled on the left hand side and we have \[57+6\] on the right hand side (RHS) of the equality.
We now have,
\[\Rightarrow 7{{x}^{2}}=57+6=63\]
That is,
\[\Rightarrow 7{{x}^{2}}=63\]
Now, we will divide both sides by 7, since we are dividing by the entity on both the side of the equality, therefore the equality is not changed, we get,
\[\Rightarrow \dfrac{7{{x}^{2}}}{7}=\dfrac{63}{7}\]
Solving which, we will have the expression in terms of \[{{x}^{2}}\], that is,
\[\Rightarrow {{x}^{2}}=9\]
Which can also be written as,
\[\Rightarrow {{x}^{2}}={{(3)}^{2}}\]
Removing the square on both sides, we will have the value of x, which is, \[x=3\]
But since \[{{x}^{2}}=9\]
So the value of x can also be -3.
Therefore, \[x=3,-3\]
Note:
In quadratic equations, we often encounter expressions like \[{{x}^{2}}=9\], at first it may seem x has only one value but need to keep in mind that: the number of values of x which can be obtained from a polynomial equation is directly proportional to the degree of that polynomial expression.
So, \[{{x}^{2}}=9\], has the value of not only \[x=3\] but also \[x=-3\],
As \[{{(3)}^{2}}={{(-3)}^{2}}=9\].
Complete step by step solution:
According to the given question, we have to solve in terms of x. The given polynomial expression is of the degree 3 and so it is a quadratic equation. The given equation will have two values which when substituted in the equation will satisfy the condition.
We will start first with rearranging the equation, we will separate the x components and the constant terms and then proceed with solving for x.
We have,
\[7{{x}^{2}}-6=57\]
On the left hand side (LHS) of the equality, we have a negative number 6, so we in order to separate the x components and the constant terms, we will add 6 on both the sides of the equality, we get
\[\Rightarrow (7{{x}^{2}}-6)+6=57+6\]
So, the 6 will get cancelled on the left hand side and we have \[57+6\] on the right hand side (RHS) of the equality.
We now have,
\[\Rightarrow 7{{x}^{2}}=57+6=63\]
That is,
\[\Rightarrow 7{{x}^{2}}=63\]
Now, we will divide both sides by 7, since we are dividing by the entity on both the side of the equality, therefore the equality is not changed, we get,
\[\Rightarrow \dfrac{7{{x}^{2}}}{7}=\dfrac{63}{7}\]
Solving which, we will have the expression in terms of \[{{x}^{2}}\], that is,
\[\Rightarrow {{x}^{2}}=9\]
Which can also be written as,
\[\Rightarrow {{x}^{2}}={{(3)}^{2}}\]
Removing the square on both sides, we will have the value of x, which is, \[x=3\]
But since \[{{x}^{2}}=9\]
So the value of x can also be -3.
Therefore, \[x=3,-3\]
Note:
In quadratic equations, we often encounter expressions like \[{{x}^{2}}=9\], at first it may seem x has only one value but need to keep in mind that: the number of values of x which can be obtained from a polynomial equation is directly proportional to the degree of that polynomial expression.
So, \[{{x}^{2}}=9\], has the value of not only \[x=3\] but also \[x=-3\],
As \[{{(3)}^{2}}={{(-3)}^{2}}=9\].
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