Oops! Floor + Quadratic?

Algebra Level 5

For real numbers a a and b b , the equations x 2 − 3 x + 4 = 0 x^2-3x+4=0 and 4 x 2 − 2 ⌊ 3 a + b ⌋ x + b = 0 4x^2-2\lfloor3a+b\rfloor x+b=0 have a common root. What is the complete set of values of a a ?

[ − 11 3 , − 10 3 ) \left [\frac{-11}{3},\frac{-10}{3}\right ) [ − 10 3 , − 3 ] \left [\frac{-10}{3},-3\right ] ( − 11 3 , − 10 3 ) \left (\frac{-11}{3},\frac{-10}{3}\right ) [ − 10 3 , − 3 ) \left [\frac{-10}{3},-3\right ) [ − 11 3 , − 10 3 ] \left [\frac{-11}{3},\frac{-10}{3}\right ] None of the other options ( − 10 3 , − 3 ) \left (\frac{-10}{3},-3\right )

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Pranjal Jain by Pranjal Jain , 23, India

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1 solution

Pranjal Jain
Jan 15, 2015

Discriminant of x 2 − 3 x + 4 x^2-3x+4 comes out to be 3 2 − 4 × 1 × 4 = − 7 3^2-4\times 1\times 4=-7 . So the equation x 2 − 3 x + 4 x^2-3x+4 has two complex conjugate roots! Now as the coefficients in other equation are real, they also must be complex conjugates. In a nutshell, these equations must have both roots common.

1 4 = − 3 − 2 [ 3 a + b ] = 4 b \frac{1}{4}=\frac{-3}{-2[3a+b]}=\frac{4}{b}

⇒ b = 16 \Rightarrow b=16

⇒ [ 3 a + 16 ] = 6 \Rightarrow [3a+16]=6

⇒ [ 3 a ] = − 10 \Rightarrow [3a]=-10

⇒ − 10 ≤ 3 a < − 9 \Rightarrow -10\leq 3a< -9

⇒ − 10 3 ≤ a < − 3 \Rightarrow \dfrac{-10}{3}\leq a< -3

10 Helpful 0 Interesting 0 Brilliant 0 Confused

the main point to note is that both roots are common

Samarth Agarwal - 5 years, 8 months ago

@Pranjal Jain you have missed the negative sign in the last line ,

Rudraksh Sisodia - 5 years ago

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Corrected. Thanks.

Pranjal Jain - 5 years ago

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