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3 years ago

Which of the following are solutions to the equation below? Check all that apply. 4x2 - 12x + 9 = 5

Mathematics

1 answer:

Anika [276]3 years ago

4 0

Step-by-step explanation:

4x² - 12x + 9 = 5 → 4x² - 12x + 4 = 0 → x² - 3x + 1 = 0

∆ = b² - 4ac → ∆ = (-3)² -4(1)(1) = 9 - 4 = 5

x = (-(-3) ± √5)/2(1) = (3 ± √5)/2

x² - 3x + 1 = 0 → (x - 3/2)² - 9/4 + 1 = 0 → (x - 3/2)² = 5/4 → x - 3/2 = ±√5 / 2 → x = ½( 3 ± √5)

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csec%5Cleft%28x%5Cright%29%7D%7B%5Ccos%5Cleft%28x%5Cright%29%7D-%5Cfrac%7B%5Csin%5

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Step-by-step explanation:

First, convert all the secants and cosecants to cosine and sine, respectively. Recall that $csc(x)=1/sin(x)$ and $sec(x)=1/cos(x)$ .

Thus:

$\frac{sec(x)}{cos(x)} -\frac{sin(x)}{csc(x)cos^2(x)}$

$=\frac{\frac{1}{cos(x)} }{cos(x)} -\frac{sin(x)}{\frac{1}{sin(x)}cos^2(x) }$

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$\frac{\frac{1}{cos(x)} }{cos(x)}=\frac{1}{cos(x)} \cdot \frac{1}{cos(x)}=\frac{1}{cos^2(x)}$

For the second term:

$\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}$

So, all together: (same denominator; combine terms)

$\frac{1}{cos^2(x)}-\frac{sin^2(x)}{cos^2(x)}=\frac{1-sin^2(x)}{cos^2(x)}$

Note the numerator; it can be derived from the Pythagorean Identity:

$sin^2(x)+cos^2(x)=1; cos^2(x)=1-sin^2(x)$

Thus, we can substitute the numerator:

$\frac{1-sin^2(x)}{cos^2(x)}=\frac{cos^2(x)}{cos^2(x)}=1$

Everything simplifies to 1.

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3 years ago

Which of the following are solutions to the equation below? Check all that apply. 4x2 - 12x + 9 = 5​

Which of the following are solutions to the equation below? Check all that apply. 4x2 - 12x + 9 = 5