Difference between revisions of "1955 AHSME Problems/Problem 12"
Awesomechoco (talk | contribs) (→Solution 1) |
Megaboy6679 (talk | contribs) |
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<cmath>2\cdot\sqrt{5x^2-6x+1}+6x-2-6x=4-6x \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}-2 =-6x+4</cmath> | <cmath>2\cdot\sqrt{5x^2-6x+1}+6x-2-6x=4-6x \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}-2 =-6x+4</cmath> | ||
− | |||
After that, adding <math>2</math> to both sides will give us | After that, adding <math>2</math> to both sides will give us | ||
<cmath>2\cdot\sqrt{5x^2-6x+1}-2+2=-6x+4+2 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}=-6x+6</cmath> | <cmath>2\cdot\sqrt{5x^2-6x+1}-2+2=-6x+4+2 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}=-6x+6</cmath> | ||
+ | Next, we divide both sides by 2 which gives us | ||
+ | |||
+ | <cmath>\frac{2\cdot\sqrt{5x^2-6x+1}-2+2}{2}=\frac{-6x+4+2}{2} \Longrightarrow \sqrt{5x^2-6x+1}=-3x+3</cmath> | ||
+ | Finally, solving the equation, we get | ||
+ | |||
+ | <cmath>5x^2-6x+1=(-3x+3)^2 \Longrightarrow 5x^2-6x+1=9x^2-18x+9</cmath> | ||
+ | <cmath>\Longrightarrow 5x^2-6x+1-(9x^2-18x+9)=9x^2-18x+9-(9x^2-18x+9) </cmath> | ||
+ | <cmath>\Longrightarrow -4x^2+12x-8=0 \Longrightarrow -4(x-1)(x-2)=0</cmath> | ||
+ | <cmath>\Longrightarrow x-1=0 \text{or}\ x-2=0 \Longrightarrow x=1 \text{or}\ x=2</cmath> | ||
+ | Plugging 1 and 2 into the original equation, <math>\sqrt{5x-1}+\sqrt{x-1}=2</math>, we see that when <math>x=1</math> | ||
+ | |||
+ | <cmath>\sqrt{5x-1}+\sqrt{x-1}=2 \Longrightarrow \sqrt{5\cdot1-1}+\sqrt{1-1}=2 \Longrightarrow \sqrt4+\sqrt0=2 \Longrightarrow 2+0=2 \Longrightarrow 2=2</cmath> | ||
+ | the equation is true. On the other hand, we note that when <math>x=2</math> | ||
+ | |||
+ | <cmath>\sqrt{5x-1}+\sqrt{x-1}=2 \Longrightarrow \sqrt{5\cdot2-1}+\sqrt{2-1}=2 \Longrightarrow \sqrt9+\sqrt1=2 \Longrightarrow 3+1=2 \Longrightarrow 4=0</cmath> | ||
+ | the equation is false. | ||
+ | Therefore the answer is <math>\boxed{{\textbf{(D) }} x=1}</math>. | ||
− | + | ~awesomechoco | |
+ | |||
+ | ==Solution 2== | ||
+ | Let us test the answer choices, for it is in this case simpler and quicker. <math>x=0</math> and <math>x=2/3</math> obviously doesn't work, since square roots of negative numbers are not real. <math>x=2</math> doesn't work either, because the first term is already bigger than 2, and the second term cannot be negative. By the process of elimination, the answer is <math>\boxed{{\textbf{(D) }} x=1}</math>. | ||
+ | |||
+ | ~megaboy6679 | ||
+ | |||
+ | == See Also == | ||
+ | {{AHSME box|year=1955|num-b=11|num-a=13}} | ||
− | + | {{MAA Notice}} |
Latest revision as of 20:48, 31 January 2023
Contents
Problem
The solution of is:
Solution 1
First, square both sides. This gives us
Then, adding
to both sides gives us
After that, adding
to both sides will give us
Next, we divide both sides by 2 which gives us
Finally, solving the equation, we get
Plugging 1 and 2 into the original equation,
, we see that when
the equation is true. On the other hand, we note that when
the equation is false.
Therefore the answer is
.
~awesomechoco
Solution 2
Let us test the answer choices, for it is in this case simpler and quicker. and
obviously doesn't work, since square roots of negative numbers are not real.
doesn't work either, because the first term is already bigger than 2, and the second term cannot be negative. By the process of elimination, the answer is
.
~megaboy6679
See Also
1955 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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