# Math Help - Quadratic Equation Part 1

1. ## Quadratic Equation Part 1

How to do these?

1. Find the real roots of $2x^2+3x-1=0$. Give your answer in surd form.

2. Find the range of values of q if $x^2+(q+1)x+q+1=0$ has complex roots.

3. Find the range of values of m if $x^2+x+1=m(x+2)$ has real and distinct roots.

4. Find the range of values of k if $x^2+(k-3)x+k=0$ has roots with same sign.

1. $\frac{1}{2}(-3 \pm \sqrt{17})$

2. -1<q<3

3. $m<-3-2\sqrt{3},m>-3+2\sqrt{3}$

4. k>0

2. For (4), you got part of it correct (namely the product of the roots is positive). But you also want your roots to be real, since it doesn’t make any sense to talk about two conjugate complex numbers having the same sign.

3. 1. Find the real roots of $2x^2+3x-1=0$. Give your answer in surd form.
\begin{aligned}
x &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} \\
&= \frac{-3 \pm \sqrt{9 + 8}}{4} \\
&= \frac{-3 \pm \sqrt{17}}{4} \\
&= \frac{1}{4}(-3 \pm \sqrt{17})
\end{aligned}

Hmm, I'm getting a different answer.

2. Find the range of values of q if $x^2+(q+1)x+q+1=0$ has complex roots.
A quadratic equation in the form $ax^2 + bx + c = 0$ has complex roots if the discriminant $b^2 - 4ac < 0$. So
a = 1
b = q + 1
c = q + 1
\begin{aligned}
b^2 - 4ac &< 0 \\
(q + 1)^2 - 4(1)(q + 1) &< 0 \\
q^2 + 2q + 1 - 4q - 4 &< 0 \\
q^2 - 2q - 3 &< 0 \\
(q - 3)(q + 1) &< 0
\end{aligned}

Make a sign chart. Draw a number line and the critical points -1 and 3 (you get these by setting each factor equal to 0).
Code:
----+----+----+----+----+----+----+----
-1                   3
You have 3 intervals to test:
(-∞, -1)
(-1, 3)
(3, ∞)

From each interval, pick a number inside it to test into the inequality (q - 3)(q + 1) < 0, in order to determine the sign. The signs of the polynomial for each interval is as follows:
Code:
----+----+----+----+----+----+----+----
pos    -1        neg        3 pos
So the answer is -1 < q < 3.

01

4. Originally Posted by yeongil
\begin{aligned}
x &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} \\
&= \frac{-3 \pm \sqrt{9 + 8}}{4} \\
&= \frac{-3 \pm \sqrt{17}}{4} \\
&= \frac{1}{4}(-3 \pm \sqrt{17})
\end{aligned}

Hmm, I'm getting a different answer.
I find this too...

5. 3. Find the range of values of m if $x^2+x+1=m(x+2)$ has real and distinct roots.
This is the same as #2, but this time, since we want real and distinct roots, the discriminant has to be positive, or $b^2 - 4ac > 0$. So
\begin{aligned}
x^2 + x + 1 &= m(x + 2) \\
x^2 + x + 1 &= mx + 2m \\
x^2 + x - mx + 1 - 2m &= 0 \\
x^2 + (1 - m)x + (1 - 2m) &= 0 \\
\end{aligned}

a = 1
b = 1 - m
c = 1 - 2m
\begin{aligned}
b^2 - 4ac &> 0 \\
(1 - m)^2 - 4(1)(1 - 2m) &> 0 \\
1 - 2m + m^2 - 4 + 8m &> 0 \\
m^2 + 6m - 3 &> 0
\end{aligned}

I want to find the zeros of the polynomial $m^2 + 6m - 3$. Let's pretend for a moment that it equals 0 and solve for m:
\begin{aligned}
m^2 + 6m - 3 &= 0 \\
m^2 + 6m &= 3 \\
m^2 + 6m + 9 &= 3 + 9 \\
(m + 3)^2 &= 12 \\
m + 3 &= \pm\sqrt{12} \\
m &= -3 \pm 2\sqrt{3} \\
\end{aligned}

Make a sign chart. Draw a number line and the critical points (the zeros we found earlier):
Code:
---------+-------------------+---------
-6.46                0.46
( $0.46 \approx -3 + 2\sqrt{3}$ and $-6.46 \approx -3 - 2\sqrt{3}$)

You have 3 intervals to test:
$(-\infty,\;-3 - 2\sqrt{3})$
$(-3 - 2\sqrt{3},\;-3 + 2\sqrt{3})$
$(-3 + 2\sqrt{3}, \infty)$

From each interval, pick a number inside it to test into the inequality $m^2 + 6m - 3 > 0$, in order to determine the sign. The signs of the polynomial for each interval is as follows:
Code:
---------+-------------------+---------
pos -6.46      neg       0.46 pos
So the answer is $m<-3-2\sqrt{3},m>-3+2\sqrt{3}$.

01

6. Originally Posted by yeongil
\begin{aligned}
! I got that answer too! But the answer in my book is $\frac{1}{2}$... Maybe there is a mistake in my book?
7. Yeah. Looks like a mistake. I got $\frac{1}{4}(-3\pm\sqrt{17})$ too