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,
 $$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}\to \left(3\right)$$
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 $$x$$
 $$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}\to \left(3\right)$$
By substituting these values of $$a,b$$ , and $$c$$ in the equation $$\to \left(3\right)$$
 $$x={\displaystyle \frac{-\left(-7\right)\pm \sqrt{{\left(-7\right)}^2-4\times 1\times 12}}{2\times 1}}$$
By solving the above equation, we get
 $$x={\displaystyle \frac{+7\pm \sqrt{49-48}}{2}}$$
Let subtract the two terms inside the root we get,
 $$x={\displaystyle \frac{+7\pm \sqrt{1}}{2}}$$
We know that $$1$$ can also be written as $$1^2$$ so that square and root cancelled each other we get $$1$$
 $$x={\displaystyle \frac{7+1}{2}}$$
Due to $$\pm$$ we get two values for $$x$$
Case: $$1$$
 $$x={\displaystyle \frac{7+1}{2}}$$
 $$x={\displaystyle \frac{8}{2}}$$
 $$x=4$$
Case: $$2$$
 $$x={\displaystyle \frac{7-1}{2}}$$
 $$x={\displaystyle \frac{6}{2}}$$
 $$x=3$$
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 $$1$$ .