# Thread: Polynomials questions again... part 5

1. ## Polynomials questions again... part 5

I don't know how to solve this! Can someone show the formula?

1. Find all possibles values of k such that the equation $x^2-(k-3)x+k^2+2k+5=0$ has real roots. If $\alpha$ and $\beta$ are real roots of this equation, show that $\alpha^2+\beta^2=-(k+5)^2+24$. Hence, find the maximum value $\alpha^2+\beta^2$.

2a. If $p(x)=3x^2+5x-2$ is a factor of the polynomial $q(x)=3x^4+8x^3+kx^2+3x-2$, determine the value of k. With this value of k, find all four zeroes of $q(x)$.

2b. Find the solution set of the inequality $x[p(x)]>0$.

Answer:
1. $\{k:-\frac{11}{3}\leq k \leq -1\}$; Maximum value $= \frac {200}{9}$

2a. $k=6; -2, \frac{1}{3}, -\frac{1}{2}\pm \frac{\sqrt{3}}{2}i$

2b. $\{x:-2 or $x>\frac{1}{3}\}$

2. Originally Posted by cloud5
I don't know how to solve this!

1. Find all possibles values of k such that the equation $x^2-(k-3)x+k^2+2k+5=0$ has real roots. If $\alpha$ and $\beta$ are real roots of this equation, show that $\alpha^2+\beta^2=-(k+5)^2+24$. Hence, find the maximum value $\alpha^2+\beta^2$.

2a. If $p(x)=3x^2+5x-2$ is a factor of the polynomial $q(x)=3x^4+8x^3+kx^2+3x-2$, determine the value of k. With this value of k, find all four zeroes of $q(x)$.

2b. Find the solution set of the inequality $x[p(x)]>0$.

Answer:
1. $\{k:-\frac{11}{3}\leq k \leq -1\}$; Maximum value $= \frac {200}{9}$

2a. $k=6; -2, \frac{1}{3}, -\frac{1}{2}\pm \frac{\sqrt{3}}{2}i$

2b. $\{x:-2 or $x>\frac{1}{3}\}$
For number 1 note that it's a quadratic in x so use the discriminant (b^2-4ac) and set it greater than 0 to solve for real roots

3. Originally Posted by cloud5
I don't know how to solve this!

1. Find all possibles values of k such that the equation $x^2-(k-3)x+k^2+2k+5=0$ has real roots. If $\alpha$ and $\beta$ are real roots of this equation, show that $\alpha^2+\beta^2=-(k+5)^2+24$. Hence, find the maximum value $\alpha^2+\beta^2$.

2a. If $p(x)=3x^2+5x-2$ is a factor of the polynomial $q(x)=3x^4+8x^3+kx^2+3x-2$, determine the value of k. With this value of k, find all four zeroes of $q(x)$.

2b. Find the solution set of the inequality $x[p(x)]>0$.

Answer:
1. $\{k:-\frac{11}{3}\leq k \leq -1\}$; Maximum value $= \frac {200}{9}$

2a. $k=6; -2, \frac{1}{3}, -\frac{1}{2}\pm \frac{\sqrt{3}}{2}i$

2b. $\{x:-2 or $x>\frac{1}{3}\}$
Q1. Start by using the fact that you require the discriminant of the quadratic to be greater than zero.

Q2 a. Start by noting that $3x^2 + 5x - 2 = (3x - 1)(x + 2)$. Therefore $q(-2) = 0$.

Q2b. Start by taking a graphical approach and sketch the graph of $y = x p(x)$. For what values of $x$ is $y > 0$?