Okay, this polynomial is crazy.

Algebra Level 2

Consider the polynomial $p(x) = ax^2 + bx + c$ has no real roots. Provided that $25a+5b+c>0$ , which of the following condition must be true?

c = 0 Can't be determined c > 0 c < 0

3 solutions

Pranav Rao
Dec 10, 2015

The graph of the polynomial lies above the x axis for all values of x as the discriminant is negative and the value of the polynomial at x=5 is positive. So at x=0 the polynomial is c which should be positive.

Rohit Shah
Mar 28, 2014

Use continuity

I wonder if the answer is D instead of B

Hsu Je Jang - 5 years, 2 months ago

Anuran Sonowal
Apr 19, 2014

for roots to be complex, b^2-4ac<0 here a=25, b=5 so b^2-4ac= 25-100c this quantity will be negative iff c>0

This is wrong. There is nothing in the problem that says $a=25$ and $b=5$ . The statement that $25a+5b+c>0$ means that p(5)>0. From here the conclusion is that $a>0$ . The discriminant $b^2-4ac$ is negative, which together with $a>0$ indeed implies $c>0$ as well.

Luís Sequeira - 5 years, 6 months ago

I agree, the question is badly written

Jake Williams - 5 years, 5 months ago

Eu concordo amiguinho haha

Nicholas Yamasaki - 5 years, 5 months ago

How are you concluding that a>0? Can't a be something like -1, and b be say 5, and c be > 0 and the equation 25a+5b+c>0 will still be true

Adéolá Adélékè - 5 years, 3 months ago

Since p(x) has no real roots, its graph never crosses the x-axis. It lies entirely above or below it. Since p(5) > 0, p(x) must lie entirely above the x-axis. Thus, p(x) must open upward, otherwise its graph would intersect the x-axis. Thus, a > 0.

Also, since p(x) has no real roots, its discriminant is negative. Thus, b^2 - 4ac < 0. Thus, 0 < b^2 < 4ac. Thus, either a and c are both positive or both negative. So a cannot be negative.

Scott Bartholomew - 2 years, 11 months ago

Okay, this polynomial is crazy.

3 solutions

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