# 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 \(a\neq 0\).

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## Solving by Factoring

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 so that \( ax^2 + bx + c = a(x+p)(x+q) \). By the zero product property: \[ \text{if}\quad ax^2 + bx + c = a(x+p)(x+q) = 0 ,\quad \\\text{then either}\quad x = -p \text{ or } x = -q \]

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

\(1)\) We have to break \(b\) (Coefficient of \(x\)) into 2 two terms in such a way that their sum\(=b\) and their product\(=ac\) \[\begin{equation}ax^2+bx+c=0 \\ ax^2+(b_1+b_2)x+c=0 \text{ such that }b_1+b_2=b \text{ and } b_1 \times b_2=ac \end{equation}\]

\(2)\) Next we need to group \(ax^2 + b_1x \text{ and } b_2x+c\) and factorise 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. \((x-\alpha)=0 \implies x=\alpha, (x-\beta)=0 \implies x=\beta\).

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

Following the steps mentioned above, we first break the coefficient of \(x\) in two terms such that their sum\(=5\) and their product\(=1 \times 6=6\) \[x^2+(2+3)x +6=0 \text{ Here we can observe that }2+3=5 \text{ and } 2 \times 3=6. \\ x^2+2x+3x+6=0 \\ x(x+2)+3(x+2)=0 \\ \text{ Taking out } (x+2) \text{ as a common factor }, (x+3)(x+2)=0 \\ x+3=0 \implies x=-3 \text{ and } x+2=0 \implies x=-2\]

Therefore the two roots of this equation are \(\boxed{-3,-2}\).

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 \(0\). (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.

*We have \[ x^2 - x - 6 = (x-3)(x+2) .\]

Note that \( x^2 - x - 6 \) can also be expressed as \( (x^2 -x - 6) \). The factors of \( x^2 - x - 6 \) are \(1\), \(x^2 - x - 6\), \(x-3\) and \(x+2\), so we know that \( x = 3 \) or \( x = -2 \). \(_\square\)

Solve the equation \( x^2+3x+2=0 \) for \(x\).

\[\begin{align} x^2+3x+2 &=0 \\ x^2+2x+x+2 &=0\\ x(x+2)+1(x+2) & =0\\ (x+2)(x+1) & =0 \end{align}\] So either, \[(x+2)=0 \text{ or } (x+1)=0 \\ \therefore x=-2 \text{ or } x =-1 \\ x = -2, -1\]

**Note:** However, that 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.

## Finding quadratic equation from roots

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: \[(x-a)(x-b)=0\\x^2 -(a+b)x+ab= 0\]

Find the quadratic equation whose roots are \(2\) and \(-3\)

Solution:Lets consider the equation in variable \(x\), we have the following:

\[x=2\implies (x-2)=0\\ x=-3\implies (x+3)=0\]Multiplying both the equations:

\[\begin{align} (x-2)(x+3) & =0\\ x(x+3)-2(x+3) & =0\\ x^2+3x-2x-6 & =0\\ x^2+x-6 & =0 \end{align}\]

Find the quadratic equation whose roots are \(5\) and \(6\)

Solution:Lets consider the equation in variable \(x\), we have the following:

\[x=5 \implies x-5=0\\ x=6\implies x-6=0\]Multiplying both the equations:

\[\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 \end{align}\]

## Solving by Completing the Square

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 \( x^2 + 8x + 10 \).

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:

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

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

Solution:First take \(2\) as common \(2\left(x^{2}+\dfrac{3x}{2}\right)+1=0\). Since our middle term is \(\dfrac 32 x\) 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 whole equation as

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

Thus we have: \[x+\dfrac{3}{4} =\pm \dfrac{1}{4}\\ \boxed{x =\left(\dfrac{-1}{2} , -1\right)}\]

## Solving by Quadratic Formula

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 for \(x\): \(5x^2-2x-3=0\)

Solution:Here, \(a=5\) ; \(b=-2\) ; \(c=-3\). Using quadratic formula:\[\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}\\ \therefore x & = (-0.6,1) \end{align}\]

Solve for \(x\): \(x^2-4x+1\)

Solution:Here, \(a=1\) ; \(b=-4\) ; \(c=1\). Using quadratic formula:\[\begin{align} x & = \dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}\\ & = \dfrac {-(-4) \pm \sqrt{[-4]^2 - 4×1×1}}{2×1}\\ & = \dfrac {4 \pm \sqrt{16 -4}}{2} =\dfrac {4 \pm \sqrt{12}}{2}\\ &=\dfrac {4 \pm 2×\sqrt {3}}{2}\\ \therefore x & =(2+\sqrt3 , 2-\sqrt3) \end{align}\]

Solve for x:- \(x^2-20x-69\) = \(0\)

\(a=1\); \(b=-20\); \(c=-69\).

Substituting values in quadratic formula, we get:-

=\[\begin {align} x & = \dfrac {-(-20) \pm \sqrt {(-20)^2 -4×1×-69}}{2×1}\\ & = \dfrac {20 \pm \sqrt {400 +276}}{2}\\ & = \dfrac {20 \pm \sqrt {676}}{2}\\ & = \dfrac {20 \pm 26}{2}\\ \therefore x & = (23,-3) \end{align}\]

Try the following problem.

Using the quadratic formula to find the roots of a quadratic equation i.e. \( x= \dfrac { -b\pm \sqrt { b ^2 - 4ac } }{ 2a } \), find the roots of the equation \[ x^2−20x−69=0.\]

## Parabolas

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 \(\left(\dfrac {-1}{2} ,2 \right)\).

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

\[\begin{align} 2 &= A \times\left (\dfrac {-1}{2}\right) ^2 \\ A &= 8\\ y &= 8x^2 \end{align}\]

Try the following problems.

If the McDonald's logo were stored as a set of pixels, enlargement would quickly result in distorted or pixelated images, which are an eyesore. As such, companies often make vector images of their logos, in which the information is stored as mathematical formulae. Such vector images are easily scaled while maintaining sharp, crisp images.

As a first approximation, the logo is deconstructed and approximated as 2 parabolic curves of the form \( y = -A(x-5)^2 \) and \( y = - A (x+5)^2 \). The McDonald's logo has a height to length ratio of 1.05. What is \(A\)?

## Nature of roots of quadratic equation

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 lead 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 \(ax^2+bx+c=0\). Discriminant, \(D\), for the Quadratic Formula is \(b^2-4ac\).

\[D=b^2-4ac\]

\[\begin{cases} b^2-4ac \gt 0 & \text{Two distinct real roots} \\ b^2-4ac=0 & \text{Equal and real roots} \\ b^2-4ac \lt 0 & \text{Imaginary roots} \end{cases}\]

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

\[\begin{align} 1.\quad 2x^2+x-1&=0 \\ 2. \quad x^2-4x+4 &=0 \end{align}\]

\(1.\quad 2x^2+x-1 \quad\text{ here } a=2,b=1,c=-1\)

\[\begin{align} \therefore b^2-4ac & =1^2-4 \times 2 \times -1\\ & =9 \\ \because b^2-4ac & \gt 0, \quad\text{The roots are real and distinct} \end{align}\]

\(2.\quad x^2-4x+4=0 \text{ here }a=1,b=-4,c=4 \)

\[\begin{align} \therefore b^2-4ac & =(-4)^2-4 \times 1 \times 4\\ & =0 \\ \because b^2-4ac & =0, \text{The roots are real and equal} \end{align}\]

Find the value of \(k\) for which the the given quadratic equation has equal roots.

\[x^2+4x+k\]

We know that if \(D=0\) then the equation has equal roots. So,

\[b^2-4ac=0\\ (4)^2-4(1)(k)=0\\ \boxed{k=4}\]

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

Solution:Here \(a=1\), \(b=d\) and \(c=-1\). So the discriminant would be:\[D=d^2-4×1×-1=d^2+4\]

Now in this case as \(d^2\) is a perfect square, it is always greater than or equal to \(0\). So:

\[d^2+4\geq 4\]

Thus, the discriminant is always greater than \(0\). So, this equation distinct roots for any value of \(d\).

## Word Problems - Basic

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

Let the present age of son be \(x\) years. So, son's age two years ago was \(x-2\) years and therefore, his father's age two years ago was \(3×(x-2)^2\). This implies the present age of father us \([3×(x-2)^2]+2\), hence after three years his age will be \([3×(x-2)^2] +2+ 3\)= \([3×(x-2)^2] +5\). And son's age 3 years hence will be \(x+3\).

According to given conditions, \[ \begin{array}{} 3(x-2)^2+5 & = 4(x+3) \\ \Rightarrow 3x^2-16x+5 & =0 \\ \Rightarrow (3x-1)(x-5) & =0 \\ \Rightarrow x & =\frac 13, 5. \end{array} \] So, either \(3x-1=0\) or \(x-5=0\)

So, either \(x\) = \(\Large \frac {1}{3}\) or \(x=5\)

But, if we use, \(x\) = \(\Large \frac {1}{3}\), then age of the son before 2 years would go in negative and age cannot be negative.

So, \(x=5\).

So, the present age of the son is \(5\) years.

The present age of the man is \([3×(x-2)^2]+2\) = \([3×(5-2)^2]+2\)= \([3×3^2]+2\) = \([3×9]+2\) = \(27+2\) = \(29\) years. \(_\square\)

Find two numbers whose sum is \(40\) and product \(375\).

Let one number be \(x\). Then, according to first condition, the second number should be \(40-x\). Substituting, the values of the numbers in the second condition, we get:

\[\begin{align} x(40-x) &= 375\\ 40x-x^2 &= 375\\ x^2-40x+375 &=0\\ x^2-25x-15x+375 &=0\\ x(x-25)-15(x-25) &=0\\ (x-25)(x-15) &=0\ \end{align}\]

Therefore, we can the have: \[\begin{align} x-25=0 &\implies x=25\quad\text{or}\\ x-15 =0 &\implies x=15\\ \therefore\text{ if } x=25, &\quad 40-x=15\\ \text{and if,}\quad x=15, &\quad 40-x = 25 \end{align}\]

So, larger number is \(25\) and smaller number is \(15\).

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

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

\[ n^2 +n -90 = (n - 9)(n+10) =0, \]

and thus \( n = 9 \). Then the two numbers are 9 and 10, and their sum is 19.

Try the following problems.

The difference of the cubes of two consecutive odd positive integers is 400 more than the sum of their squares. Find the sum of the two integers.

**Clarification**: The odd positive integers are \( 1, 3, 5, 7, 9, \ldots.\) Two consecutive odd positive integers refer to two consecutive numbers in this sequence. It does not refer to two consecutive integers (of which one will not be odd).

## Biquadratic equations

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

Solve \( x^{4} - 3x^{2} + 1 = 0 \).

Solution: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 \( u = x^{2} = \dfrac{3 \pm \sqrt{5}}{2} \), solving that equation for x gives:

\[ \boxed{\pm\sqrt{\dfrac{3 \pm \sqrt{5}}{2}}} \]

## Quadratic Equations - Problem Solving

This section contains miscellaneous problems on quadratic equations to try. And eventually your problem solving skills will be enhanced.

## See Also

**Cite as:**Quadratic Equations.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/quadratic-equations/