Get out of here with your quadratic!

Algebra Level 2

True or False?

\quad If x < 7 x<7 , then x 2 > 7 x x^2>7x can never be correct.

True False

Pi Han Goh by Pi Han Goh , 30, Malaysia

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3 solutions

Hung Woei Neoh
Jul 15, 2016

x 2 > 7 x x 2 − 7 x > 0 x ( x − 7 ) > 0 x < 0 , x > 7 x^2>7x\\ x^2-7x>0\\ x(x-7)>0\\ x<0,\;x>7

Notice that x < 0 x<0 satisfies the above inequality, and x < 0 x<0 satisfies x < 7 x<7 as well.

Therefore, x 2 > 7 x x^2>7x can be correct. The answer is False \boxed{\text{False}}

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Romain Milon
Jul 15, 2016

You probably should precise the nature of x x (which I supposed to be a real number). In the case if this problem, we only have to find a value of x x that satisfies the equation: − 4 -4 .

− 4 < 7 -4 < 7 and ( − 4 ) 2 > 7 × ( − 4 ) < = > 16 > − 28 (-4)^2 > 7\times(-4) <=> 16 > -28

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Tom Regan
Aug 5, 2016

I'm no math major, but wouldn't any negative integer satisfy this equation?

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Nope, it is possible that the inequality holds true. Set x = − 10 x = -10 (a negative integer), then x 2 = 100 x^2 = 100 and 7 x = − 70 7x = -70 , so x 2 > 7 x x^2 > 7x can be true.

Why? Notice that for x > 7 x>7 , if we multiply the inequality by x x , should we reverse the sign of the inequality? Why or why not?

Pi Han Goh - 4 years, 10 months ago

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