Question

How do you solve \[{x^2} - 7x + 12 = 0\] using the quadratic formula \[?\]
A. \[x\] = \[4\] or \[3\]
B. \[x\] = \[5\] or \[6\]
C. \[x\] = \[1\] or \[2\]
D. \[x\] = \[0\] or \[5\]

Answer 1

Hint: The given question is describing the operation of using quadratic formula, addition/subtraction/multiplication/division. Also, remind the quadratic formula and compare the given equation with the quadratic formula to solve the given equation.

Complete step-by-step answer:
The given equation is shown below,
 \[{x^2} - 7x + 12 = 0 \to \left( 1 \right)\]
We know that the quadratic formula is,
 \[a{x^2} + bx + c = 0 \to \left( 2 \right)\]
Then,
 \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \to \left( 3 \right)\]
Let compare equation \[\left( 1 \right)\] and \[\left( 2 \right)\]
For finding the value of \[a,b\] and \[c\] in the given equation
 \[{x^2} - 7x + 12 = 0\] \[ \to \left( 1 \right)\]
 \[a{x^2} + bx + c = 0 \to \left( 2 \right)\]
So let compare the \[{x^2}\] terms in equations \[\left( 1 \right)\] and \[\left( 2 \right)\]
 \[1 \times {x^2}\]
 \[a \times {x^2}\]
 So we find the value of \[a\] that is
 \[a = 1\]
Let compare the \[x\] terms in equations \[\left( 1 \right)\] and \[\left( 2 \right)\]
 \[ - 7 \times x\]
 \[b \times x\]
So we find the value of \[b\] , which is
 \[b = - 7\]
Let compare the constant terms in the equation \[\left( 1 \right)\] and \[\left( 2 \right)\]
 \[12\]
 \[c\]
So we find the values of \[c\] that is
 \[c = 12\]
So we get \[a,b\] , and \[c\] values are \[1, - 7\] , and \[12\] respectively.
Let substitute these values in the equation \[\left( 3 \right)\] for finding the values of \[x\]
 \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \to \left( 3 \right)\]
By substituting these values of \[a,b\] , and \[c\] in the equation \[ \to \left( 3 \right)\]
 \[x = \dfrac{{ - \left( { - 7} \right) \pm \sqrt {{{\left( { - 7} \right)}^2} - 4 \times 1 \times 12} }}{{2 \times 1}}\]
By solving the above equation, we get
 \[x = \dfrac{{ + 7 \pm \sqrt {49 - 48} }}{2}\]
Let subtract the two terms inside the root we get,
 \[x = \dfrac{{ + 7 \pm \sqrt 1 }}{2}\]
We know that \[1\] can also be written as \[{1^2}\] so that square and root cancelled each other we get \[1\]
 \[x = \dfrac{{7 + 1}}{2}\]
Due to \[ \pm \] we get two values for \[x\]
Case: \[1\]
 \[x = \dfrac{{7 + 1}}{2}\]
 \[x = \dfrac{8}{2}\]
 \[x = 4\]
Case: \[2\]
 \[x = \dfrac{{7 - 1}}{2}\]
 \[x = \dfrac{6}{2}\]
 \[x = 3\]
In, case: \[1\] we assume \[ \pm \] it as a “ \[ + \] ”.
In, case: \[2\] we assume \[ \pm \] as a “ \[ - \] “.
By considering the \[ \pm \] as “ \[ + \] ” we get the \[x\] value is \[4\] . By considering the \[ \pm \] as “ \[ - \] “we get \[x\] value as \[3\] .
So the final value \[x\] is
 \[x = 4\] or \[x = 3\]
So, the correct answer is “Option A”.

Note: To find the value \[x\] of from the given equation we would compare the equation with the quadratic formula. After comparing the equation we would find the value of \[a,b\] , and \[c\] . When substituting these values in quadratic formula remind the following things
1) When a negative number is multiplied by the negative number the answer becomes a positive number.
2) When a positive number is multiplied with another positive number the answer becomes a positive number.
3) In multiplication any one term is negative the answer becomes negative.
The square root value \[1\] is always \[1\] .