How do you solve $64 x^{2} - 144 x + 81 = 0$ $64x^2 - 144x + 81 = 0$ ?

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How do you solve $64 x^{2} - 144 x + 81 = 0$ $64x^2 - 144x + 81 = 0$ ?

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Answer 1 · 2015-05-18T00:18:08

We can use Bhaskara here.

$\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$

where $a = 64$ $a=64$ , $b = - 144$ $b=-144$ and $c = 81$ $c=81$ .

$\frac{144 \pm \sqrt{144^{2} - 4 (64) (81)}}{128}$ $(144+-sqrt(144^2-4(64)(81)))/128$

$\frac{144 \pm \sqrt{20736 - 20736}}{128}$ $(144+-sqrt(20736-20736))/128$

$\frac{144 \pm \sqrt{0}}{128}$ $(144+-sqrt(0))/128$

$\frac{144}{128} = \frac{9}{8}$ $144/128=color(green)(9/8)$

Your quadratic function is tangent to axis $x$ $x$ , touching it only in the point with $x$ $x$ coordinate $(\frac{9}{8})$ $(9/8)$ . So, we can say that $x_{1} = x_{2} = \frac{9}{8}$ $x_1=x_2=9/8$

Answer 2 · 2015-05-18T00:20:30

$y = {(8 x - 9)}^{2} = 64 x^{2} - 144 x + 81 .$ $y = (8x - 9)^2 = 64x^2 - 144x + 81.$ = 0, then:

$(8 x - 9) = 0 \to x = \frac{9}{8} .$ $(8x - 9) = 0 -> x = 9/8.$ There is double root.

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How do you solve $64 x^{2} - 144 x + 81 = 0$ $64x^2 - 144x + 81 = 0$ ?

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