Question

How do you solve using the quadratic formula ${{x}^{2}}-4x+10=0$?

Answer 1

In this problem, we have to solve and find the value of x using the quadratic formula. We can compare the general formula of the quadratic equation and the given quadratic equation to find the value of a, b, c. We can then substitute those values of a, b, c in the quadratic formula to find the value of x.

Complete step by step solution:
We know that the given quadratic equation is,
${{x}^{2}}-4x+10=0$ ….. (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 = -4, 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
$\Rightarrow x=\dfrac{-\left( -4 \right)\pm \sqrt{{{\left( -4 \right)}^{2}}-4\times 1\times \left( 10 \right)}}{2\times 1}$
Now we can simplify the above step, we get

& \Rightarrow x=\dfrac{\left( 4 \right)\pm \sqrt{16-40}}{2\times 1} \\\ & \Rightarrow x=\dfrac{4\pm \sqrt{4\left( -6 \right)}}{2} \\\ & \Rightarrow x=2\pm \sqrt{-6} \\\ \end{aligned}$$ Now we can separate the terms to simplify it, $$\begin{aligned} & \Rightarrow x=2+\sqrt{-6} \\\ & \Rightarrow x=2-\sqrt{-6} \\\ \end{aligned}$$ **We can see that this equation has no real solutions, but we are allowed complex solutions. Therefore, the value of $$x=2-i\sqrt{6}$$ and $$x=2+i\sqrt{6}$$.** **Note:** We can also use a simple factorisation method to solve this problem but we have to know that the answer is in complex form, so we can solve only by using a quadratic formula to get the final answer correctly. Students may make mistakes in the quadratic formula part, which should also concentrate while substituting the values in the quadratic formula.