Question

$\pink{ \huge \mathbb{Q} \sf{uestion}}$
Without solving comment upon the nature of roots of each of the following equation:

$\sf {6x}^{2} - 13x + 4 = 0$

♨️No spam $\red\checkmark$
♨️no plagiarism[tex]\pink\checkmark[/tex
-_-​

Answer 1

$\huge\ \pmb{\red{«\: คꈤ \mathfrak Sฬєя \: » }}$

$6x ^ { 2 } -13x+4$

• Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1} and \: x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$

$6x^{2}-13x+4=0$

• All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula $: \frac{-b±\sqrt{b^{2}-4ac}}{2a}$The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 6\times 4}}{2\times 6} \\ \\ x=\frac{-\left(-13\right)±\sqrt{169-4\times 6\times 4}}{2\times 6} \\ \\ x=\frac{-\left(-13\right)±\sqrt{169-24\times 4}}{2\times 6} \\ \\ x=\frac{-\left(-13\right)±\sqrt{169-24\times 4}}{2\times 6} \\ \\ x=\frac{-\left(-13\right)±\sqrt{169-96}}{2\times 6} \\ \\ x=\frac{-\left(-13\right)±\sqrt{73}}{2\times 6} \\ \\ x=\frac{13±\sqrt{73}}{2\times 6} \\ \\ x=\frac{13±\sqrt{73}}{12} \\ \ \\ x=\frac{\sqrt{73}+13}{12} \\ \\ x=\frac{13-\sqrt{73}}{12}$

• Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$.

• Substitute $\frac{13+\sqrt{73}}{12} \: for \: x_{1} \: and \: \frac{13-\sqrt{73}}{12} \: for \: x_{2}.$

$6x^{2}-13x+4=6\left(x-\frac{\sqrt{73}+13}{12}\right)\left(x-\frac{13-\sqrt{73}}{12}\right)$

$\: \: \: \: \: \: \: \: \: \: \huge \bold{@WildCat7083 }$

Answer 2

Step-by-step explanation:

• (6x)2−13x+4=0
• Step 1: Simplify both sides of the equation.

36x2−13x+4=0

Step 2: Find discriminant with a=36, b=-13, c=4.

b2−4ac

=(−13)2−4(36)(4)

=−407

Answer:

−407