Question

How do you solve ${{x}^{2}}-10x+18=0$ using the quadratic formula?

Answer 1

In this problem we need to solve the given quadratic equation i.e., we need to calculate the values of $x$ where the given equation is satisfied. For solving a quadratic equation, we have several methods. But in the problem, they have mentioned to use the quadratic formula which is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Now we will compare the given equation with the standard quadratic equation $a{{x}^{2}}+bx+c=0$ and write the values of $a$, $b$, $c$. Now we will substitute those values in the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and simplify the obtained equation to get the required result.

Complete step by step solution:
Given equation ${{x}^{2}}-10x+18=0$.
Comparing the above quadratic equation with standard quadratic equation $a{{x}^{2}}+bx+c=0$, then we will get the values of $a$, $b$, $c$ as
$a=1$, $b=-10$, $c=18$.
We have the quadratic formula for the solution as
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values of $a$, $b$, $c$ in the above equation, then we will get
$\Rightarrow x=\dfrac{-\left( -10 \right)\pm \sqrt{{{\left( -10 \right)}^{2}}-4\left( 1 \right)\left( 18 \right)}}{2\left( 1 \right)}$
We know that when we multiplied a negative sign with the negative sign, then we will get positive sign. Applying the above rule and simplifying the above equation, then we will get
$\begin{aligned} & \Rightarrow x=\dfrac{10\pm \sqrt{100-72}}{2} \\\ & \Rightarrow x=\dfrac{10\pm \sqrt{28}}{2} \\\ \end{aligned}$
In the above equation we have the value $\sqrt{28}$. We can write the number $28$ as $4\times 7$. Now the value of $\sqrt{28}$ will be $\sqrt{28}=\sqrt{{{2}^{2}}\times 7}=2\sqrt{7}$. Substituting this value in the above equation, then we will get
$\Rightarrow x=\dfrac{10\pm 2\sqrt{7}}{2}$
Taking $2$ as common and simplifying the above equation, then we will get
$\begin{aligned} & \Rightarrow x=\dfrac{2\left( 5\pm \sqrt{7} \right)}{2} \\\ & \Rightarrow x=5\pm \sqrt{7} \\\ \end{aligned}$
Hence the solution of the given quadratic equation ${{x}^{2}}-10x+18=0$ is $5\pm \sqrt{7}$.

Note: We can also check whether the obtained solution is correct or wrong by substituting either $x=5+\sqrt{7}$ or $x=5-\sqrt{7}$ in the given equation. Substituting $x=5+\sqrt{7}$ in the given equation ${{x}^{2}}-10x+18=0$, then we will get
${{\left( 5+\sqrt{7} \right)}^{2}}-10\left( 5+\sqrt{7} \right)+18=0$
Simplifying the above equation using ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, then we will get
$\begin{aligned} & \Rightarrow {{5}^{2}}+{{\left( \sqrt{7} \right)}^{2}}+2\left( 5 \right)\left( \sqrt{7} \right)-10\times 5-10\sqrt{7}+18=0 \\\ & \Rightarrow 25+7+10\sqrt{7}-50-10\sqrt{7}+18=0 \\\ \end{aligned}$
The term $10\sqrt{7}-10\sqrt{7}$ will become zero, then we will get
$\begin{aligned} & \Rightarrow 50-50=0 \\\ & \Rightarrow 0=0 \\\ & \Rightarrow LHS=RHS \\\ \end{aligned}$
Hence the obtained solution is correct.