Which of the following are solutions to the equation below?4x^2-12x+9=5

Mathematics

2 answers:

timofeeve [1]3 years ago

8 0

Answer:

$x=\frac{-\sqrt{5}+3}{2}$ and $x=\frac{\sqrt{5}+3}{2}$

Step-by-step explanation:

$4x^2-12x+9=5$

To solve for x we need to make right hand side 0

Subtract 5 on both sides

$4x^2-12x+4=0$

Divide whole equation by 4

$x^2-3x+1=0$

Apply quadratic formula to solve for x

$x=\frac{-b+-\sqrt{b^2-4ac}}{2a}$

the value of a=1, b=-3 and c=1

$x=\frac{-(-3)+-\sqrt{(-3)^2-4(1)(1)}}{2(1)}$

$x=\frac{3+-\sqrt{5}}{2}$

$x=\frac{-\sqrt{5}+3}{2}$ and $x=\frac{\sqrt{5}+3}{2}$

Damm [24]3 years ago

5 0

$4x^2-12x+9=5\\4x^2-12x+9-5=0\\4x^2-12x+4=0 \ \ \ |:4\\x^2-3x+1=0 \\ \\ x= \cfrac{3б \sqrt{(-3)^2-4} }{2}= \cfrac{3б \sqrt{9-4} }{2}= \cfrac{3б \sqrt{5} }{2} 

$

Correct answers are C and E

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Cos(θ) < 0, so we know it would be in Quadrant 2 or 3
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==> sin(θ) = $\frac{1}{257}$
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sin(θ) = opp./hyp both of opp and hyp are positive but adj suppose to negative because that way it leads the cos(θ) < 0
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