Question 6
Find the smallest integer value of for which is always positive.

Ask by Mullins Mann. in Singapore

Mar 31,2025

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Question 6 Find the smallest integer value of $$ k $$ for which $$ 5 x^{2}-7 x+11+k $$ is always positive.

UpStudy AI · Accepted Answer

We want the quadratic $$5x^2-7x+11+k$$ to be positive for all values of $$x$$. Since the coefficient of $$x^2$$ is positive ($$5 > 0$$), the quadratic is always positive if its discriminant is negative. 1. Write the quadratic in standard form: $$ 5x^2-7x+(11+k). $$ 2. Compute the discriminant $$\Delta$$ using: $$ \Delta = b^2 - 4ac, $$ where $$a = 5$$, $$b = -7$$, and $$c = 11+k$$. Thus, $$ \Delta = (-7)^2 - 4(5)(11+k)= 49 - 20(11+k). $$ 3. Simplify the discriminant: $$ \Delta = 49 - 220 - 20k = -171 - 20k. $$ 4. Set the discriminant to be negative: $$ -171 - 20k < 0. $$ 5. Solve the inequality: $$ -20k < 171 \quad \Rightarrow \quad k > -\frac{171}{20}. $$ Simplify: $$ -\frac{171}{20} = -8.55. $$ 6. Since $$k$$ must be an integer, the smallest integer greater than $$-8.55$$ is: $$ k = -8. $$ Thus, the smallest integer value of $$k$$ for which $$5x^2-7x+11+k$$ is always positive is $$\boxed{-8}$$. The smallest integer value of $$ k $$ is $$-8$$.

Question 6
Find the smallest integer value of for which is always positive.

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Question 6 Find the smallest integer value of for which is always positive.