So much constraints!

Algebra Level 4

x 2 − ( 2 k − 3 ) x + ( 2 k − 4 ) = 0 x^2 - (2k-3)x + (2k-4) = 0

What is the smallest possible integer k k such that the equation above have a solution which is greater than 3 3 and the other solution smaller than 3 3 ?


The answer is 4.

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Zhaochen Xie by Zhaochen Xie , 19, Spain

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2 solutions

Saad Mukarram
Mar 29, 2015

2 Helpful 0 Interesting 0 Brilliant 0 Confused

For every value of k, (2k-5)^2 is greater than 0

Harshit Singhania - 6 years ago

k>2.5 is sufficient in order to have 2 solutions. But, finding the correct answer using "plugging" can be a long process if the required k is big. I suggest solving the quadratic equation using the quadratic formula and than you get 2 solutinos: x 1 = 2 k − 4 , x 2 = 1 x 1 > 3 ⇒ 2 k − 4 > 3 ⇒ k > 3.5 ⇒ k = 4 \\ x_{1}=2k-4 \: , \: x_{2}=1 \\ x_{1}>3 \: \: \Rightarrow \: \: 2k-4>3 \: \: \Rightarrow \\ k>3.5 \; \; \; \Rightarrow \: \: \boxed{k=4}

Noa Samoray - 4 years, 11 months ago
Venkata Teja B
Mar 29, 2015

let a,b be two roots

a+b=2k-3

ab=2k-4

a+b+3=ab+4

a=1 for b other than one

k=(b+4)/2. b>3

for k to be integer,b=4

therefore k=4

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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