Quadratic equation (exercise)

Type classification: this is a quiz resource.

Exercise 1

Solve for $x\in ℝ$: $x^2-5x+6=0$

Solution:

The equation is of the form $a{x}^2+bx+c=0$ where $a=1$, $b=-5$ and $c=6$.

$\mathrm{\Delta }=(-5{)}^2-4\times 1\times 6=25-24=1$

Note that $\mathrm{\Delta }$ is positive. There are thus two solutions:

$x_{1,2}=\frac{-b\pm \sqrt{\mathrm{\Delta }}}{2a}=\frac{-(-5)\pm \sqrt{1}}{2}=\frac{5\pm 1}{2}$

The two solutions are $x_1=\frac{5-1}{2}=2$ and $x_2=\frac{5+1}{2}=3$

Exercise 2

Solve for $x\in ℝ$: $2x^2-2x-12=0$

Solution:

The equation is of the form $a{x}^2+bx+c=0$ where $a=2$, $b=-2$ and $c=-12$.

$\mathrm{\Delta }=(-2{)}^2-4\times 2\times (-12)=4+96=100$

Note that $\mathrm{\Delta }$ is positive. There are thus two solutions:

$x_{1,2}=\frac{-b\pm \sqrt{\mathrm{\Delta }}}{2a}=\frac{-(-2)\pm \sqrt{100}}{2\times 2}=\frac{2\pm 10}{4}$

The two solutions are $x_1=\frac{2-10}{4}=-2$ and $x_2=\frac{2+10}{4}=3$

Exercise 3

Solve for $x\in ℝ$: $x^2-4x+4=0$

Solution:

The equation is of the form $a{x}^2+bx+c=0$ where $a=1$, $b=-4$ and $c=4$.

$\mathrm{\Delta }=(-4{)}^2-4\times 1\times 4=16-16=0$

Since $\mathrm{\Delta }$ is zero, there is only a single solution:

$x=\frac{-b}{2a}=\frac{-(-4)}{2\times 1}=\frac{4}{2}=2$

Alternatively, one can solve this equation by using the 2nd binomial formula:

$\begin{array}{cc}{x}^2-4x+4& =0\\ (x-2{)}^2& =0\\ x-2& =0\\ x& =2\\ \end{array}$

Exercise 4

Solve for $x\in ℝ$: $x^2-4x+5=0$

Solution:

The equation is of the form $a{x}^2+bx+c=0$ where $a=1$, $b=-4$ and $c=5$.

$\mathrm{\Delta }=(-4{)}^2-4\times 1\times 5=16-20=-4$

Since $\mathrm{\Delta }$ is negative, there is no real solution, $x\notin ℝ$.

Exercise 5

Solve for $x\in ℝ$: $x^2-\frac{3}{2}x+\frac{1}{2}=0$

Solution:

The equation is of the form $a{x}^2+bx+c=0$ where $a=1$, $b=-\frac{3}{2}$ and $c=\frac{1}{2}$.

$\mathrm{\Delta }={(-\frac{3}{2})}^2-4\times 1\times \frac{1}{2}=\frac{9}{4}-\frac{8}{4}=\frac{1}{4}$

Note that $\mathrm{\Delta }$ is positive. There are thus two solutions:

$x_{1,2}=\frac{-b\pm \sqrt{\mathrm{\Delta }}}{2a}=\frac{-(-\frac{3}{2})\pm \sqrt{\frac{1}{4}}}{2\times 1}=\frac{\frac{3}{2}\pm \frac{1}{2}}{2}=\frac{3\pm 1}{4}$

The two solutions are $x_1=\frac{3-1}{4}=\frac{1}{2}$ and $x_2=\frac{3+1}{4}=1$.

Note that one can avoid the fraction here by multiplying all terms of the quadratic equation by 2.

Exercise 6

Solve for $x\in ℝ$: $x^2-4x+3=0$

Solution:

The equation is of the form $a{x}^2+bx+c=0$ where $a=1$, $b=-4$ and $c=3$.

$\mathrm{\Delta }=(-4{)}^2-4\times 1\times 3=16-12=4$

Note that $\mathrm{\Delta }$ is positive. There are thus two solutions:

$x_{1,2}=\frac{-b\pm \sqrt{\mathrm{\Delta }}}{2a}=\frac{-(-4)\pm \sqrt{4}}{2}=\frac{4\pm 2}{2}$

The two solutions are $x_1=\frac{4-2}{2}=1$ and $x_2=\frac{4+2}{2}=3$

Exercise 7

Solve for $x\in ℝ$: $100x^2-700x+600=0$

Solution:

One approach is to first divide all terms of this equation by 100 to avoid unnecessarily large numbers.

$x^2-7x+6=0$

The equation is of the form $a{x}^2+bx+c=0$ where $a=1$, $b=-7$ and $c=6$.

$\mathrm{\Delta }=(-7{)}^2-4\times 1\times 6=49-24=25$

Note that $\mathrm{\Delta }$ is positive. There are thus two solutions:

$x_{1,2}=\frac{-b\pm \sqrt{\mathrm{\Delta }}}{2a}=\frac{-(-7)\pm \sqrt{25}}{2}=\frac{7\pm 5}{2}$

The two solutions are $x_1=\frac{7-5}{2}=1$ and $x_2=\frac{7+5}{2}=6$

Exercise 8

Solve for $x\in ℝ$: $x^2-x-dx+d=0$ where $d\in ℝ$.

Solution:

The equation is of the form $a{x}^2+bx+c=0$ where $a=1$, $b=-(1+d)$ and $c=d$.

$\mathrm{\Delta }=(1+d{)}^2-4\times 1\times d=1+2d+d^2-4d=1-2d+d^2=(1-d{)}^2$

Note that $\mathrm{\Delta }$ is positive (or zero if $d=1$). There are thus two (or one) solution(s):

$x_{1,2}=\frac{-b\pm \sqrt{\mathrm{\Delta }}}{2a}=\frac{1+d\pm |1-d|}{2}$

There is no need to discuss the sign of $1-d$ as both cases are handled anyway with the $\pm$ sign.

The solutions are $x_1=\frac{1+d-(1-d)}{2}=d$ and $x_2=\frac{1+d+1-d}{2}=1$.

Indeed, when $d=1$, both solutions are equal.

Exercise 9

Find the intersection between the parabola $f(x)=x^2-x-2$ and the line $g(x)=x+1$.

Solution:

At the intersection $f(x)=g(x)$:

$\begin{array}{cc}{x}^2-x-2& =x+1\\ x^2-2x-3& =0\end{array}$

The discriminant:

$\mathrm{\Delta }=(-2{)}^2-4\times 1\times (-3)=4+12=16$

Note that $\mathrm{\Delta }$ is positive. There are thus two solutions:

$x_{1,2}=\frac{-b\pm \sqrt{\mathrm{\Delta }}}{2a}=\frac{2\pm 4}{2}$

The solutions are $x_1=-1$ and $x_2=3$. The corresponding $y$-values are $y_1=x_1+1=0$ and $y_2=x_2+1=4$.

$f(x)$ and $g(x)$ intersect thus at the points (-1,0) and (3,4).

The graph of the functions $f(x)$ and $g(x)$ are illustrated below.

Exercise 10

Find the intersection between the parabola $f(x)=2x^2-3$ and the line $g(x)=2x+1$.

Solution:

At the intersection $f(x)=g(x)$:

$\begin{array}{cc}2x^2-3& =2x+1\\ 2x^2-2x-4& =0\end{array}$

The discriminant:

$\mathrm{\Delta }=(-2{)}^2-4\times 2\times (-4)=4+32=36$

Note that $\mathrm{\Delta }$ is positive. There are thus two solutions:

$x_{1,2}=\frac{-b\pm \sqrt{\mathrm{\Delta }}}{2a}=\frac{2\pm 6}{4}$

The solutions are $x_1=-1$ and $x_2=2$. The corresponding $y$-values are $y_1=2x_1+1=-1$ and $y_2=2x_2+1=5$.

$f(x)$ and $g(x)$ intersect thus at the points (-1,-1) and (2,5).

The graphs of the functions $f(x)$ and $g(x)$ are illustrated below.

Exercise 11

Find all values of $k$ so that the parabola $f(x)=x^2+3x+k$ and the line $g(x)=-x+1$ are tangent to each other. Compute the coordinates of the point of tangency.

Solution:

$f(x)$ and $g(x)$ are tangent if they intersect at a single point. The intersection can be computed by solving $f(x)=g(x)$:

$\begin{array}{cc}{x}^2+3x+k& =-x+1\\ x^2+4x+k-1& =0\end{array}$

The discriminant must be zero:

$\begin{array}{cc}\mathrm{\Delta }=(4{)}^2-4(k-1)& =0\\ 16-4(k-1)& =0\\ 4-(k-1)& =0\\ 4-k+1& =0\\ k& =5\end{array}$

The solution is $x=\frac{-b}{2a}=\frac{-4}{2}=-2$ and the corresponding $y$-value is $y=-x+1=-3$.

$f(x)$ and $g(x)$ are tangent if $k=5$ and the coordinates of the point of tangency is (-2,-3).

The curves $f(x)$ and $g(x)$ are illustrated below for k = 5.