Question

How do you solve $8{{x}^{2}}-6x+1=0$?

Answer 1

The degree of the equation is the highest power to which the variable is raised. The degree of the equation decides whether the equation is linear, quadratic, cubic, etc. If the degree of the equation is two, then it is quadratic. We can find the roots of a quadratic equation $a{{x}^{2}}+bx+c=0$ using the quadratic formula method as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

Complete step by step solution:
We are given the quadratic expression $8{{x}^{2}}-6x+1=0$. On comparing with the general solution of the quadratic equation $a{{x}^{2}}+bx+c$, we get $a=8,b=-6\And c=1$.
To solve the quadratic equation, we need to find its roots. We can find the roots of the equation using the formula method.
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substituting the values of the coefficients in the above formula, we get

& \Rightarrow x=\dfrac{-(-6)\pm \sqrt{{{\left( -6 \right)}^{2}}-4(8)(1)}}{2(8)} \\\ & \Rightarrow x=\dfrac{6\pm 2}{16} \\\ \end{aligned}$$ $$\Rightarrow x=\dfrac{6+2}{16}$$ or $$x=\dfrac{6-2}{16}$$ Simplifying the above expressions, we get $$\Rightarrow x=\dfrac{8}{16}$$ or $$x=\dfrac{4}{16}$$ **Canceling out the common factor from both numerator and denominator for the above fraction, we get the roots of the quadratic equation as $$x=\dfrac{1}{2}$$ or $$x=\dfrac{1}{4}$$.** **Note:** There are many other methods to solve a quadratic equation like the factorization method, completing the square method, hit and trial method. We can use any of them to solve. The factorization method should be preferred because it gives the value of two roots of the equation, whether they are real or not. For example, we can use this method to find the roots of the equation $${{x}^{2}}+x+1=0$$ , and can directly state that it has complex or imaginary roots, using its discriminant. It may be difficult to solve it with other methods.