X² + X -6 equal zero

1 answer:

3 0

So if we have x^2+x-6=0 we can use FOIL to solve

x^2+x-6=0
(x+3)(x-2)=0
x=-3,2

Hope this helps you understand how to solve using the FOIL method which means First, Outside,Inside,Last

Any questions please ask! Please help me out and rate my answer Brainliest if my answer helped you out!

Thank you so much!

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Nimfa-mama [501]

A = LW......W = A/L
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The perimeter of a rectangle is 50 ft. the length is 1 ft more than the width. what is the length and width

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mr_godi [17]

Answer:

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Step-by-step explanation:

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2 years ago

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :