Question

How do you solve $ {x^2} - 11x + 28 = 0$ using the quadratic formula $ ?$

Answer 1

Hint: In this question using the quadratic formula. First we take the given equation. We identify the value of $ a,b$ and $ c$ in the quadratic equation. After that Substitute the values $ a ,b$ and $ c$ into the quadratic formula and solve for $ x$ . Now simplify the equation, hence we get the equation.
Use the quadratic formula to find the solutions,
\[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Pull terms out from under the radical, assuming positive real numbers. Now we find the $ x$ value. First we separate the positive and sign values.
Finally we get $ x$ values.

Complete step-by-step solution:
The given quadratic equation is $ {x^2} - 11x + 28 = 0$ .
Use the quadratic formula to find the solutions,
\[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Substitute the values $ a = 1,b = - 11$ and $ c = 28$ into the quadratic formula and solve for $ x$
 $ \dfrac{{ - ( - 11) \pm \sqrt {{{( - 11)}^2} - 4(1 \times 28)} }}{{2 \times 1}}$
Now simplify the equation, hence we get
 $ \dfrac{{11 \pm \sqrt {{{( - 11)}^2} - 4(1 \times 28)} }}{{2 \times 1}}$
Raise $ - 11$ to the power of $ 2$
$ \Rightarrow$ $ x = \dfrac{{11 \pm \sqrt {121 - 4(1 \times 28)} }}{{2 \times 1}}$
Multiply $ 28$ by $ 1$
$ \Rightarrow$ \[x = \dfrac{{11 \pm \sqrt {121 - 4 \times 28} }}{{2 \times 1}}\]
Multiply $ - 4$ by $ 28$
$ \Rightarrow$ \[x = \dfrac{{11 \pm \sqrt {121 - 112} }}{{2 \times 1}}\]
Subtract $ 112$ from $ 121 $
$ \Rightarrow$ \[x = \dfrac{{11 \pm \sqrt 9 }}{{2 \times 1}}\]
Rewrite $ 9$ as $ {3^2}$
$ \Rightarrow$ \[x = \dfrac{{11 \pm \sqrt {{3^2}} }}{{2 \times 1}}\]
Pull terms out from under the radical, assuming positive real numbers.
$ \Rightarrow$ \[x = \dfrac{{11 \pm 3}}{{2 \times 1}}\]
Multiply $ 2$ by$ 1$
$ \Rightarrow$ \[x = \dfrac{{11 \pm 3}}{2}\]
Now we find the $ x$ value. First we separate the positive and sign values,
Let,
$ \Rightarrow$ $ x = \dfrac{{11 + 3}}{2}$
Add $ 11$ by $ 3$
$ \Rightarrow$ $ x = \dfrac{{14}}{2}$
Divide $ 14$ by $ 2$
$ \Rightarrow$ $ x = 7$
Let, $ x = \dfrac{{11 - 3}}{2}$
Subtract $ 3$ from $ 11$
$ \Rightarrow$ $ x = \dfrac{8}{2} $
Divide $ 8$ by $ 2$
$ \Rightarrow$ $ x = 4$
The given equation solution is $ 7$ and $ 4$

Therefore the values of x are 7 and 4.

Note: The formula for finding the roots of the quadratic equation $ a{x^2} + bx + c = 0$ is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
The formula for finding roots of a quadratic equation was known to ancient Babylonians, though not in a form as we derived.
They found the roots by creating the steps as a verse, which is a common practice at their times. Babylonians used quadratic equations for deciding to choose the dimensions of their land for agriculture.