Question

How do you solve using the completing the square method ${{x}^{2}}-5x+10=0$?

Answer 1

In this problem, we have to solve the given quadratic equation by completing the square method. We should know to change the given equation into a complete square equation using the algebraic whole square formula to find the missing term for the complete square equation and we can add it on both sides. We can then take the square root on both sides to cancel the square and to find the value of x.

Complete step by step solution:
We know that the given quadratic equation to be solved is,
${{x}^{2}}-5x+10=0$ ……. (1)
Now we can subtract 10 on both sides in the above equation (1), we get
${{x}^{2}}-5x=-10$….. (2)
We can now take the first two terms ${{x}^{2}}-5x$
We know that,
${{\left( x-a \right)}^{2}}={{x}^{2}}-2ax+{{a}^{2}}$
We can see that,

& \Rightarrow -2a=-5 \\\ & \Rightarrow a=\dfrac{5}{2} \\\ & \Rightarrow {{a}^{2}}={{\left( \dfrac{5}{2} \right)}^{2}}=\dfrac{25}{4} \\\ \end{aligned}$$ We can add the above value in both the left-hand side and the right-hand side of equation (2), to get a perfect square equation, we get $${{x}^{2}}-5x+{{\left( \dfrac{5}{2} \right)}^{2}}=-10+\dfrac{25}{4}$$ We can write the left-hand side in whole square form as it is a perfect square equation, we get $$\begin{aligned} & \Rightarrow {{\left( x-\dfrac{5}{2} \right)}^{2}}=\dfrac{-10\times 4+25}{4} \\\ & \Rightarrow {{\left( x-\dfrac{5}{2} \right)}^{2}}=\dfrac{-15}{4} \\\ \end{aligned}$$ Now we can take square root on both sides, we get $$\begin{aligned} & \Rightarrow {{\sqrt{\left( x-\dfrac{5}{2} \right)}}^{2}}=\sqrt{\dfrac{-15}{4}} \\\ & \Rightarrow \left( x-\dfrac{5}{2} \right)=\pm \dfrac{i\sqrt{15}}{2} \\\ \end{aligned}$$ We can see that the roots are not real, so we can make complex roots. **Therefore, the value of $$x=\dfrac{5}{2}\pm \dfrac{i\sqrt{15}}{2}$$.** **Note:** We can also use the quadratic formula to find or check the values we get are correct. We know that the given quadratic equation is, $${{x}^{2}}-5x+10=0$$ ….. (1) We also know that a quadratic equation in standard form is, $$a{{x}^{2}}+bx+c=0$$ ……. (2) We can now compare the two equations (1) and (2), we get a = 1, b = -5, c = 10. We know that the quadratic formula for the standard form $$a{{x}^{2}}+bx+c=0$$ is $$x=\dfrac{-b\pm \sqrt{{{\left( b \right)}^{2}}-4\times a\times \left( c \right)}}{2a}$$ Now we can substitute the value of a, b, c in the above formula, we get $$\Rightarrow x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\times 1\times \left( 10 \right)}}{2\times 1}$$ Now we can simplify the above step, we get $$\begin{aligned} & \Rightarrow x=\dfrac{\left( 5 \right)\pm \sqrt{25-40}}{2\times 1} \\\ & \Rightarrow x=\dfrac{5\pm \sqrt{-15}}{2} \\\ & \Rightarrow x=\dfrac{5\pm i\sqrt{15}}{2} \\\ \end{aligned}$$ Therefore, the value of $$x=\dfrac{5}{2}\pm \dfrac{i\sqrt{15}}{2}$$