2018 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 10 Easy

Suppose that real number x satisfies \sqrt{49-x^2}-\sqrt{25-x^2}=3. What is the value of \sqrt{49-x^2}+\sqrt{25-x^2}? (2018 AMC 10A Problem, Question#10)

  • A.

    8

  • B.

    \sqrt{33}+8

  • C.

    9

  • D.

    2\sqrt{10}+4

  • E.

    12

Answer:A

In order to get rid of the square roots, we multiply by the conjugate. Its value is the solution.The x^2 terms cancel nicely. \left(\sqrt{49-x^2}+\sqrt{25-x^2}\right)\left(\sqrt{49-x^2}-\sqrt{25-x^2}\right)=49-x^2-25+x^2=24.

Given that \left(\sqrt{49-x^2}-\sqrt{25-x^2}\right)=3, \left(\sqrt{49-x^2}+\sqrt{25-x^2}\right)=\frac{24}{3}=\boxed{\rm (A)~8}.

Let u=\sqrt{49-x^2}, and let v=\sqrt{25-x^2}. Then v=\sqrt{u^2-24}. Substituting, we get u-\sqrt{u^2-24}=3. Rearranging, we get u-3=\sqrt{u^2-24}. Squaring both sides and solving, we get u=\frac{11}{2} and v=\frac{11}{2}-3=\frac{5}{2}. Adding, we get that the answer is \boxed{\rm (A)~8}.

Put the equations to one side. \sqrt{49-x^2}-\sqrt{25-x^2}=3 can be changed into \sqrt{49-x^2}=\sqrt{25-x^2}+3.

We can square both sides, getting us 49-x^2=\left(25-x^2\right)+\left(3^2\right)+2\cdot 3\cdot \sqrt{25-x^2}.

That simplifies out to 15=6\sqrt{25-x^2}. Dividing both sides by 6 gets us \frac{5}{2}=\sqrt{25-x^2}.

Following that, we can square both sides again, resulting in the equation \frac{25}{4}=25-x^2. Simplifying that, we get x^2=\frac{75}{4}.

Substituting into the equation \sqrt{49-x^2}+\sqrt{25-x^2}, we get \sqrt{49-\frac{75}{4}}+\sqrt{25-\frac{75}{4}}.

Immediately, we simplify into \sqrt{\frac{121}{4}}+\sqrt{\frac{25}{4}}. The two numbers inside the square roots are simplified to be \frac{11}{2} and \frac{5}{2}, so you add them up: \frac{11}{2}+\frac{5}{2}=\boxed{\rm (A)~8}.

Draw a right triangle ABC with a hypotenuse AC of length 7 and leg AB of length x.

Draw D on BC such that AD=5. Note that BC=\sqrt{49-x^2} and BD=\sqrt{25-x^2}. Thus, from the given equation, BC-BD=DC=3. Using Law of Cosines on triangle ADC, we see that \angle ADC=120^\circ, so \angle ADB=60^\circ. Since ADB is a 30-60-90 triangle, \sqrt{25-x^2}=BD=\frac{5}{2} and \sqrt{49-x^2}=\frac{5}{2}+3=\frac{11}{2}.

Finally, \sqrt{49-x^2}+\sqrt{25-x^2}=\frac{5}{2}+\frac{11}{2}=\boxed{\rm (A)~8}.

Related Problems
Problem 8 2018 AMC 10 A
Problem 9 2018 AMC 10 A
Problem 11 2018 AMC 10 A
Problem 12 2018 AMC 10 A
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