Algebra AMC 10 Math Olympiad Theory of Equations USA Math Olympiad

Quadratic Equation Problem | AMC-10A, 2005 | Problem 10

Try this beautiful problem from Algebra based on Quadratic Equation….

Quadratic equation – AMC-10A, 2005- Problem 10

There are two values of $a$ for which the equation $4 x^{2}+a x+8 x+9=0$ has only one solution for $x$. What is the sum of those values of $a$ ?

  • \(5\)
  • \(20\)
  • \(-16\)
  • \(25\)
  • \(36\)

Key Concepts


Quadratic equation

Equal roots

Check the Answer

Answer: \(-16\)

AMC-10A (2005) Problem 10

Pre College Mathematics

Try with Hints

The given equation is $4 x^{2}+a x+8 x+9=0$

\(\Rightarrow 4 x^{2}+x(a+8)+9=0\)

comparing the above equation with \(Ax^2-Bx+C=0\) we will get \(A=4\),\(B=(a+8)\),\(C=9\)

Now for equal roots of a quadratic equation \(B^2-4Ac=0\)

Can you now finish the problem ……….

Now \(B^2-4Ac=0\) becomes

\((a+8)^2-4\times 9 \times 4=0\)

\(\Rightarrow (a+8)^2=144\)

\(\Rightarrow (a+8)=\pm 12\)

\(\Rightarrow a=+4 \) & \(-20\)

Therefore The sum of the values of \(a=-20+4=-16\)

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