Solve with completing the square method:6x²-7x+2=0 and ax²+bx+c=0

1 answer:

7 0

$6x ^2-7x+2=0\\ \\ a=6 , \ b = -7 , \ c=2 \\ \\\Delta = b^{2}-4ac = (-7)^{2}-4*6*2=49-48=1 \\ \\x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{7-\sqrt{1}}{2*6}=\frac{7-1}{12} =\frac{6}{12}= \frac{1}{ 2}\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{7+\sqrt{1}}{2*6}=\frac{7+1}{12} =\frac{8}{12}= \frac{2}{ 3}$

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Show that the equation x^4/2021 − 2021x^2 − x − 3 = 0 has at least two real roots.

andreev551 [17]

The roots of an equation are simply the x-intercepts of the equation.

See below for the proof that $\mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0}$ has at least two real roots

The equation is given as: $\mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0}$

There are several ways to show that an equation has real roots, one of these ways is by using graphs.

See attachment for the graph of $\mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0}$

Next, we count the x-intercepts of the graph (i.e. the points where the equation crosses the x-axis)

From the attached graph, we can see that $\mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0}$ crosses the x-axis at approximately <em>-2000 and 2000 </em>between the domain -2500 and 2500

This means that $\mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0}$ has at least two real roots

Read more about roots of an equation at:

brainly.com/question/12912962