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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

algebra

Equal roots

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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