When does a quadratic equation have real roots?

Algebra Level 3

k x 2 17 x + 2 k = 0 kx^2-17x+2k=0

Find the maximum integral value of k k for which the above quadratic equation has two distinct real roots.


The answer is 6.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Nihar Mahajan
Jan 16, 2016

A quadratic equation has two distinct real roots if and only if its discriminant is greater than 0 0 .

Discriminant of k x 2 17 x + 2 k = ( 17 ) 2 4 ( k ) ( 2 k ) = 289 8 k 2 kx^2-17x+2k = (-17)^2-4(k)(2k)=289-8k^2

So , 289 8 k 2 > 0 k 2 < 289 8 k 2 < 36.125 289-8k^2>0 \Rightarrow k^2<\dfrac{289}{8} \Rightarrow k^2<36.125

Since 6 2 = 36 6^2=36 , maximum required k = 6 \boxed{k=6} .

Daniel Ferreira
Jan 17, 2016

Inicialmente, devemos julgar a condição, necessária, para que a equação em questão tenha duas raízes distintas.

Ora, sabemos que isso ocorre quando o valor do discriminante é maior que zero, isto e, Δ > 0 \Delta > 0 .

Isto posto, temos que:

Δ > 0 289 8 k 2 > 0 ( 6 , 01 + k ) ( 6 , 01 k ) > 0 S = { k R 6 , 01 < k < 6 , 01 } \\ \boxed{\Delta > 0} \\ 289 - 8k^2 > 0 \\ (6,01 + k)(6,01 - k) > 0 \\ \boxed{S = \left \{ k \in \mathbb{R} | - 6,01 < k < 6,01 \right \}}

Daí, o maior valor é 6.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...