F(x)=x^2 +6x-8 Show that if the equation f(x+a)=a has no real root then a^2<4(b-a)?

Question

1896 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-08T19:09:02

# (a+17) lt 0, or a lt -17#.

Prerequisite :

The quadr. eqn. #ax^2+bx+c=0# has no real roots.

# iff b^2-4ac lt 0#.

We see that, for #f(x)=x^2+6x-8#,

#f(x+a)=a rArr (x+a)^2+6(x+a)-8=a#

#rArr x^2+(2a+6)x+(a^2+5a-8)=0#.

Since this quadr. eqn. has no real roots,

# (2a+6)^2-4(1)(a^2+5a-8) lt 0#.

#:. 4(a+3)^2-4(a^2+5a-8) lt 0#.

#:. 4{(a^2+6a+9)-(a^2+5a-8)} lt 0, i.e., #

# 4(a+17) lt 0#.

#:. (a+17) lt 0, or a lt -17#.