Quadratic Equations

A quadratic equation is a polynomial equation with degree two. In other words, it is an equation of the form $ax^2 + bx + c =0$, where $a$, $b$ and $c$ are real numbers and $a\neq 0$.

Main Article: Factoring Polynomials

We can solve quadratics using factoring and the zero product property. In general, we can rewrite a quadratic as the product of two linear factors such that $ax^2 + bx + c = a(x+p)(x+q)$. By the zero product property,

$$\text{if }\ ax^2 + bx + c = a(x+p)(x+q) = 0, \ \text{then either }\ x = -p \ \text{ or }\ x = -q.$$

Now, to factorise a quadratic equation, follow these steps.

$1)$ We have to break $b$ (the coefficient of $x$) into two terms in such a way that their sum is $b$ and their product is $ac:$

$$ax^2+bx+c=0 \implies ax^2+(b_1+b_2)x+c=0 \text{ such that }b_1+b_2=b \text{ and } b_1 \times b_2=ac.$$

$2)$ Next we need to group $ax^2 + b_1x \text{ and } b_2x+c$ and factorize them in a way that they both have one factor common.

Now we will have the equation transformed into factors. From here on, the solution is easy. We use the zero product property and equate each factor to $0$, i.e. $(x-\alpha)=0 \implies x=\alpha$ and $(x-\beta)=0 \implies x=\beta$.

Solve $x^2+5x+6=0$ for $x$ by the method of factoring.

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Following the steps mentioned above, we first break the coefficient of $x$ in two terms such that their sum equals $5$ and their product equals $1 \times 6=6$: $$x^2+(2+3)x +6=0,$$ where we can observe that $2+3=5$ and $2 \times 3=6$: $$\begin{aligned} x^2+2x+3x+6&=0 \\ x(x+2)+3(x+2)&=0. \end{aligned}$$ Taking out $(x+2)$ as a common factor, we have $$\begin{aligned} (x+3)(x+2)&=0 \\ x+3&=0 \implies x=-3 \\ x+2&=0 \implies x=-2. \end{aligned}$$ Therefore the two roots of the given equation are $-3$ and $-2$. $_\square$

Method of Solving a Quadratic Equation by Factorizing:

Step 1. Make the given equation free from fractions and radicals and put it into the standard form $ax^2+bx+c=0.$

Step 2. Factorize $ax^2+bx+c$ into two linear factors.

Step 3. Put each linear factor equal to $0$ (to apply the zero product rule).

Step 4. Solve these linear equations and get two roots of the given quadratic equation.

Solve $x^2 - x - 6 =0$ by the method of factoring.

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We have $$x^2 - x - 6 = (x-3)(x+2),$$ which gives $x = 3$ or $x = -2$. $_\square$

Note that the factors of $x^2 - x - 6$ are $1, x^2 - x - 6, x-3,$ and $x+2$.

Solve the equation $x^2+3x+2=0$ for $x$.

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We have $$\begin{aligned} x^2+3x+2 &=0 \\ x^2+2x+x+2 &=0\\ x(x+2)+1(x+2) & =0\\ (x+2)(x+1) & =0. \end{aligned}$$ So,
$$\begin{aligned} (x+2)=0 \text{ or } (x+1)=0 \\ x=-2 \text{ or } x =-1.\ _\square \end{aligned}$$

Note: We cannot always factor into linear factors using only real numbers. For some quadratics $($e.g., $x^2 + 1 )$, the linear factors require complex numbers:

$$x^2 + 1 = (x+i)(x-i) .$$

Try the following problems:

When two places of the variable are given, we have to write them of the form $\text{(variable - value = 0)}$.

To find the equation from the roots:

Step 1. If the variable $x$ is given, and two values $x=a$ and $x=b$ are given, then we have to simplify them to $$x-a=0 \quad\text{ and }\quad x-b=0.$$ Step 2. Multiplying the equations and simplifying them, we arrive at this: $$\begin{aligned} (x-a)(x-b)&=0\\ x^2 -(a+b)x+ab&= 0. \end{aligned}$$

Find the quadratic equation whose roots are $2$ and $-3$

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Considering the equation in variable $x$, we have the following:
$$\begin{aligned} x=2&\implies (x-2)=0\\ x=-3&\implies (x+3)=0. \end{aligned}$$ Multiplying both the equations, we have $$\begin{aligned} (x-2)(x+3) & =0\\ x(x+3)-2(x+3) & =0\\ x^2+3x-2x-6 & =0\\ x^2+x-6 & =0.\ _\square \end{aligned}$$

Find the quadratic equation whose roots are $5$ and $6$

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Considering the equation in variable $x$, we have the following:
$$\begin{aligned} x&=5 \implies x-5=0\\ x&=6\implies x-6=0. \end{aligned}$$ Multiplying both the equations gives \[\begin{align} (x-5)(x-6) & =0\\ x(x-6)-5(x-6) & =0\\
x^2-6x-5x+30 & =0\\
x^2-11x+30 & =0.\ _\square \end{align}\]

Main Article: Completing The Square

For a quadratic polynomial $f(x) = ax^2 + bx +c$, completing the square means finding an expression of the form

$$f(x) = a(x-b)^2 + c.$$

Complete the square for the quadratic $x^2 + 8x + 10$.

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Since our middle term is $8x$, we know that we will want a perfect square with the form $(x+4)^2$, which expands to $x^2 + 8x + 16$.

Thus, we can do the following:

$$\begin{aligned} x^2 + 8x + 10 &= x^2 + 8x + 16 - 16 + 10\\ &= (x^2 + 8x + 16) - 6 \\ &= ( x+ 4 )^2 -6. \ _\square \end{aligned}$$

Solve this equation $2x^{2}+3x+1=0$ by method of completing square.

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First take $2$ as common: $2\left(x^{2}+\frac{3x}{2}\right)+1=0$. Since our middle term is $\frac 32 x,$ we know that we will want a perfect square with the form

$$\left(x+\dfrac{3}{4}\right)^2=x^{2}+\dfrac{3x}{2}+\dfrac{9}{16}.$$

So rewrite the whole equation as

$$\begin{aligned} 2\left(x+\dfrac{3x}2 +\dfrac{9}{16}-\dfrac{9}{16}\right)+1 & =0\\ 2\left(x+\dfrac{3x}2 + \left(\dfrac{3}{4}\right)^{2}-\dfrac{9}{16}\right)+1 & =0\\ 2\left(x+\dfrac{3}{4}\right)^{2}-\dfrac{1}{8} & =0\\ \left(x+\dfrac{3}{4}\right)^{2} & =\dfrac{1}{16}. \end{aligned}$$

Thus we have

$$\begin{aligned} x+\dfrac{3}{4} &=\pm \dfrac{1}{4}\\ \Rightarrow x &=-\dfrac{1}{2} \, \text{ or }\, x=-1.\ _\square \end{aligned}$$

Main Article: Quadratic Formula

The quadratic formula states that for the equation $ax^2 + bx + c =0$, the values of $x$ are given by the following:

$$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}.$$

To see how this formula is derived via completing the square, see Quadratic Formula.

Solve $5x^2-2x-3=0$ for $x$.

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Here, $a=5, b=-2, c=-3$. Using the quadratic formula, we get

\[\begin{align} x &= \dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}\\ & = \dfrac {-(-2) \pm \sqrt{(-2)^2 - 4×5×-3}}{2×5}\\ & = \dfrac {2 \pm \sqrt{4 +60}}{10} = \dfrac {2 \pm \sqrt{64}}{10}\\
& = \dfrac {2 \pm 8}{10}\\ \Rightarrow x & = -0.6 \, \text{ or }\, x=1.\ _\square \end{align}\]

Solve $x^2-4x+1$ for $x$.

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Here, $a=1, b=-4, c=1$. Using the quadratic formula, we get

$$\begin{aligned} x & = \dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}\\ & = \dfrac {-(-4) \pm \sqrt{(-4)^2 - 4×1×1}}{2×1}\\ & = \dfrac {4 \pm \sqrt{16 -4}}{2} =\dfrac {4 \pm \sqrt{12}}{2}\\ &=\dfrac {4 \pm 2×\sqrt {3}}{2}\\ \Rightarrow x & =2+\sqrt3 \, \text{ or }\, x=2-\sqrt3.\ _\square \end{aligned}$$

Solve $x^2-20x-69 = 0$ for $x.$

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Substituting the values $a=1, b=-20, c=-69$ in the quadratic formula, we get

$$\begin{aligned} x & = \dfrac {-(-20) \pm \sqrt {(-20)^2 -4×1×-69}}{2×1}\\ & = \dfrac {20 \pm \sqrt {400 +276}}{2}\\ & = \dfrac {20 \pm \sqrt {676}}{2}\\ & = \dfrac {20 \pm 26}{2}\\ \Rightarrow x & = 23 \, \text{ or }\, x=-3.\ _\square \end{aligned}$$

Try the following problems:

Main Article: Parabolas

Here is an example illustrating the above.

Find the equation of a parabola with vertex at $(0,0)$ if its axis of symmetry is the $y$-axis and its graph contains the point $\left(\frac {-1}{2}, 2 \right)$.

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We write the vertex form of the parabola as $y = A(x^2)$. Plug in the coordinates of the given point to find $A$

$$\begin{aligned} 2 &= A \times\left (\dfrac {-1}{2}\right) ^2 \\ A &= 8\\ \Rightarrow y &= 8x^2.\ _\square \end{aligned}$$

Try the following problems:

The nature of roots of a quadratic equation can be determined by observing the quadratic formula closely. It basically consists of a discriminant which actually makes the difference in formula and leads us two roots.

We know the quadratic formula is

$$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$

for any quadratic equation written in standard form of $ax^2+bx+c=0$. The discriminant $D$ for the quadratic equation is

$$D=b^2-4ac,$$

where

$$\begin{cases} b^2-4ac \gt 0: & \text{two distinct real roots} \\ b^2-4ac=0: & \text{equal and real roots} \\ b^2-4ac \lt 0: & \text{imaginary roots}. \end{cases}$$

Determine the nature of roots of the following two quadratic equations:

$$\begin{aligned} 2x^2+x-1&=0 \\ x^2-4x+4 &=0. \end{aligned}$$

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1. For the quadratic equation $2x^2+x-1=0$:
   Since $a=2,b=1,c=-1,$ $$\begin{aligned} b^2-4ac & =1^2-4 \times 2 \times -1\\ & =9 >0, \end{aligned}$$ which implies that the roots are real and distinct.

2. For the quadratic equation $x^2-4x+4=0$:
   Since $a=1,b=-4,c=4,$ $$\begin{aligned} b^2-4ac & =(-4)^2-4 \times 1 \times 4\\ & =0, \end{aligned}$$ which implies that the roots are real and repeated. $_\square$

Find the value of $k$ for which the following quadratic polynomial has repeated roots:

$$x^2+4x+k.$$

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We know that if $D=0,$ then the quadratic polynomial has repeated roots. So,

$$\begin{aligned} b^2-4ac&=0\\ (4)^2-4(1)(k)&=0\\ k&=4.\ _\square \end{aligned}$$

Show that the equation $x^2+dx-1=0$ has real and distinct roots for all real values of $d$.

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Here, $a=1, b=d,$ and $c=-1$. So the discriminant would be

$$D=d^2-4×1×-1=d^2+4.$$

Since $d^2$ is a perfect square, it is always greater than or equal to $0$. So,

$$D=d^2+4\geq 4.$$

Thus, the discriminant is always greater than $0$, implying that this equation has distinct real roots for any real value of $d$. $_\square$

Two years ago, a man's age was three times the square of his son's age. In three years, his age will be four times his son's age. Find their present ages.

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Let the present age of the son be $x$. Then the son's age two years ago was $x-2$, and his father's age two years ago was $3\times (x-2)^2$. This implies the present age of the father is $\big[3×(x-2)^2\big]+2$, and hence in three years his age will be $\big[3×(x-2)^2\big] +2+ 3= \big[3×(x-2)^2\big] +5$. The son's age in 3 years will be $x+3$.

According to given conditions, the following holds:

$$\begin{array}{rl} 3(x-2)^2+5 & = 4(x+3) \\ \Rightarrow 3x^2-16x+5 & =0 \\ \Rightarrow (3x-1)(x-5) & =0 \\ \Rightarrow x & =\frac 13, 5. \end{array}$$

If $x$ = $\frac {1}{3}$, then the son's age 2 years ago would become negative, which is impossible. So, the son's present age is $x=5$, which implies that the present age of the man is

$$\begin{aligned} 3\times (x-2)^2+2 &= 3\times (5-2)^2+2\\ &= 3\times 3^2+2\\ &= 3×9+2\\ &= 27+2\\ &= 29.\ _\square \end{aligned}$$

Find two numbers whose sum is $40$, and product $375$.

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Let one number be $x$. Then, according to the first condition, the second number is $40-x$. Substituting, the value in the second condition, we get

$$\begin{aligned} x(40-x) &= 375\\ 40x-x^2 &= 375\\ x^2-40x+375 &=0\\ x^2-25x-15x+375 &=0\\ x(x-25)-15(x-25) &=0\\ (x-15)(x-25) &=0\\ \Rightarrow x&=15, x=25. \end{aligned}$$

Therefore, the smaller number is $15$ and the larger one is $25$. $_\square$

The product of two consecutive positive integers is 90. What is their sum?

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Since the integers are consecutive, we can rewrite the expression above as $n(n+1) = 90$. This gives us the following quadratic equation: $n^2 +n -90 = 0$. Factoring, we can see that

$$n^2 +n -90 = (n - 9)(n+10) =0,$$

which implies $n = 9$. Then the two numbers are 9 and 10, and their sum is 19. $_\square$

Try the following problems:

Sometimes, the quadratic formula could be useful in solving equations of larger degree.

Solve $x^{4} - 3x^{2} + 1 = 0$.

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That equation isn't something you'd want to factor. So, you could make the substitution $u = x^{2}$. Then the equation would read

$$u^{2} - 3u + 1 = 0.$$

We can solve that with the quadratic formula:

$$u = \dfrac{3 \pm \sqrt{5}}{2}.$$

But we're not done yet. We want $x$, not $u$. Since $u = x^{2} = \frac{3 \pm \sqrt{5}}{2}$, solving that equation for $x$ gives

$$x=\pm\sqrt{\dfrac{3 \pm \sqrt{5}}{2}}.\ _\square$$

This section contains miscellaneous problems on quadratic equations for you to try, which will eventually enhance your problem solving skills.

• Factoring Polynomials
• Completing The Square
• Quadratic Formula