Question

How do you solve ${{x}^{2}}-4x+7=0$ by completing the square?

Answer 1

Assume the expression ${{x}^{2}}-4x+7=y$ and compare it with the general form given as: - $y=a{{x}^{2}}+bx+c$. Find the respective values of a, b and c. Now, find the discriminant of the given quadratic equation by using the formula: - $D={{b}^{2}}-4ac$, where D = discriminant. Now, write the expression as: - $y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]$ and substitute it equal to 0 to find the two values of x. If the value of D is negative then the roots will be of the form of complex numbers.

Complete step by step solution:
Here, we have been provided with the quadratic equation: - ${{x}^{2}}-4x+7=0$ and we are asked to solve it. That means we have to find the values of x. We have been asked to use completing the square method.
Now, we know that any quadratic equation of the form $y=a{{x}^{2}}+bx+c$ can be simplified as $y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]$, using completing the square method. Here, ‘D’ denotes the discriminant. So, on assuming ${{x}^{2}}-4x+7=y$ and comparing it with the general quadratic equation, we get,
$\Rightarrow$ a = 1, b = -4, c = 7
Applying the formula for discriminant of a quadratic equation given as, $D={{b}^{2}}-4ac$, we get,

& \Rightarrow D={{\left( -4 \right)}^{2}}-4\left( 1 \right)\left( 7 \right) \\\ & \Rightarrow D=16-28 \\\ & \Rightarrow D=-12 \\\ \end{aligned}$$ As we can see that the value of D is negative it means that we will not get any real roots but we need to find the complex roots. Therefore, substituting the values in the simplified form of y, we get, $$\begin{aligned} & \Rightarrow y=1\left[ {{\left( x+\left( \dfrac{-4}{2\times 1} \right) \right)}^{2}}-\dfrac{\left( -12 \right)}{4\times {{\left( 1 \right)}^{2}}} \right] \\\ & \Rightarrow y=\left[ {{\left( x-2 \right)}^{2}}+3 \right] \\\ \end{aligned}$$ Substituting y = 0, we get, $$\Rightarrow \left[ {{\left( x-2 \right)}^{2}}+3 \right]=0$$ $$\begin{aligned} & \Rightarrow {{\left( x-2 \right)}^{2}}+3=0 \\\ & \Rightarrow {{\left( x-2 \right)}^{2}}=-3 \\\ \end{aligned}$$ Taking square root both the sides, we get, $$\begin{aligned} & \Rightarrow \left( x-2 \right)=\pm \sqrt{-3} \\\ & \Rightarrow x-2=\pm \sqrt{3}i \\\ & \Rightarrow x=2\pm \sqrt{3}i \\\ \end{aligned}$$ Considering the two signs one – by – one, we get, $$\Rightarrow x=\left( 2+\sqrt{3}i \right)$$ or $$x=\left( 2-\sqrt{3}i \right)$$ **Hence, the solutions of the given quadratic equation are: $$x=\left( 2+\sqrt{3}i \right)$$ and $$x=\left( 2-\sqrt{3}i \right)$$.** **Note:** One may note that here we are not getting any real roots of the given equation because the value of the discriminant is less than 0. You must remember the conditions for getting the real roots and imaginary roots of a quadratic equation. One may also use the quadratic formula to solve the question and check if we are getting the same values of x. Note that the general expression of this completing the square method is also known as the vertex form. Generally, this form is used in coordinate geometry of parabola to find the vertex of the parabola which is given as $$\left( \dfrac{-b}{2a},\dfrac{-D}{4a} \right)$$. Note that the quadratic formula: - $$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$ is derived from completing the square method.