Equations Reducible to Quadratic Equations: Definition, Types, Examples

Equations Reducible to Quadratic Equations: In this article, we will discuss the equations that are not quadratic but are reducible to quadratic equations. These equations can be easily solved if we convert them to a quadratic form which is otherwise difficult to solve. Also, we shall discuss how the cubic equation is reducible to the quadratic equation to solve them.

Consider an equation: \(x^4-5x^2+12=0\). So, check that the equation can be reduced to quadratic form, then apply a simple substitution to convert it to a quadratic equation. We solve the new equation for y, where \(y=x^2\), the substitution variable, and then utilise these solutions and the substitution definition to find the real-world solutions to the equation. In most circumstances, all that is required to determine whether it is reducible to quadratic form is to check that one of the exponents is twice the other. There is one exception to this, which we’ll see when we look at some cases. Let us learn the different types of equations that can be reduced to quadratic form with solved examples.

Quadratic Equation

An equation of the form \(a x^{2}+b x+c=0\) is called a quadratic equation where \(a, b, c\) are real numbers and \(a \neq 0\).

Quadratic equations are also called polynomial equations of degree two in one variable of the form \(f(x)=a x^{2}+b x+c=0\) where \(a, b, c, \in R\) and \(a \neq 0\).

Coefficient \(a\) is called the leading coefficient as it is the coefficient of the highest degree, and \(c\) is the absolute term or the constant term of \(f(x)\). A quadratic polynomial, when equated to zero, becomes a quadratic equation. The quadratic equation will have two roots. The values of \(x\) that satisfy the equation are called the roots of the equation, namely \((\alpha, \beta)\). The roots of the quadratic equation may be either real or imaginary.

Quadratic Equation Examples

1. \(3 x^{2}-5 x+4=0\)
2. \(x^{2}+4 x+3=0\)
3. 3x(x+8)=-2
4. x(2x+3)=12

Types of Quadratic Equation

Quadratics are typically used in three ways:

1. Standard Form: y = ax^2 + bx + c
2. Factored Form: y = a(x-a)(x-b)
3. Vertex Form: y = a((x-h)^2) + k

Equations Reducible to Quadratic Equation

The primary method is to check if the equations are reduceable to quadratic form and then use the substitution method to convert them into quadratic equations. We solve the new equation to find the variable’s value and then substitute it back to the original equation to get the solution we want.

In most cases, check the exponents of the given equation. It is probably reducible to quadratic form if it is twice the other. The exceptions will be seen in the examples below.

However, these problems can be done without substitution in many cases.

Learn Concepts of Quadratic Equations

Various Types of Equations Reducible to Quadratic Equations

Type-I: Equations of the form \(a x^{2 n}+b x^{n}+c=0\)

Method: In this type of equation, we put \(x^{n}=y\), so that \(y^{2}=\left(x^{n}\right)^{2}=x^{2 n}\) and the given equation becomes \(a y^{2}+b y+c=0\), which is a quadratic equation.

Example: \(x^{\frac{2}{3}}-x^{\frac{1}{3}}+8=0\)

Putting \(x^{\frac{1}{3}}=y\) we get,

\(y^{2}-y+8=0\)

Type-II: Equations of the form \(a x+\frac{b}{x}=c, x \neq 0\)

Method: In this type of equation, we take LCM of the denominators so that the given equation becomes \(\frac{{a{x^2} + b}}{x} = x \Rightarrow a{x^2} + b = cx\)

\(\Rightarrow a x^{2}-c x+b=0\) is in the form of a quadratic equation.

Example: \(2 x+\frac{7}{x}=3, x \neq 0\)
Reducing the equation by taking the LCM of denominators, we get:

\(\frac{2 x^{2}+7}{x}=3 \Rightarrow 2 x^{2}+7=3 x\)

\(\Rightarrow 2 x^{2}-3 x+7=0\)

Type-III– Reciprocal equations

Equation of the form \(p\left(x^{2}+\frac{1}{x^{2}}\right) \pm q\left(x \pm \frac{1}{x}\right)+r=0\), where \(p, q, r\) are constants.

Method: In this type of equation, we put \(\left(x \pm \frac{1}{x}\right)=y\) then given equation becomes \(p\)
\(\left(y^{2} \pm 2\right) \pm q y+r=0\)

Example: \(3\left(x^{2}+\frac{1}{x^{2}}\right)-16\left(x+\frac{1}{x}\right)+26=0, x \neq 0\)
Substitute \(x+\frac{1}{x}=y \Rightarrow\left(x+\frac{1}{x}\right)^{2}=y^{2}\)
\(\Rightarrow x^{2}+\frac{1}{x^{2}}+2 \times x \times \frac{1}{x}=y^{2}\)
\(\Rightarrow x^{2}+\frac{1}{x^{2}}+2=y^{2}\)
\(\Rightarrow x^{2}+\frac{1}{x^{2}}=y^{2}-2\)

Substituting these values in the given equation, we get:
\(3\left(y^{2}-2\right)-16 y+26=0\)
\(\Rightarrow 3 y^{2}-6-16 y+26=0\)
\(\Rightarrow 3 y^{2}-16 y+20=0\)

Type IV: Equation of the form \(p m^{2 x}+q m^{x}+r=0\)

Method: In this type of equation, we put \(m^{x}=y\), then the given equation becomes \(a y^{2}+b y+c=0\) which is quadratic.

Example: \(9^{x+2}-6 \times 3^{x+1}+1=0\)

The given equation can be written as:
\(\left(3^{2}\right)^{x+2}-6 \times 3^{x} \times 3^{1}+1=0\)
\(\Rightarrow 3^{2 x+4}-18 \times 3^{x}+1=0\)
\(\Rightarrow 3^{2 x} \times 3^{4}-18 \times 3^{x}+1=0\)
\(\Rightarrow 81 \times 3^{2 x}-18 \times 3^{x}+1=0\)

Let \(3^{x}=y\) and square on both sides \(3^{2 x}=y^{2}\), we get:
\(81 y^{2}-18 y+1=0\)

Type V: Equation of the form \((x+a)(x+b)(x+c)(x+d)+k=0\), where \(a, b, c, d\) and \(k\) are real constants.

Method: In this type of equation, we group two terms and expand to have a similarity in the terms. The example will be the best explanation for this type of equation.

Example: \((x+1)(x+2)(x+3)(x+4)+1=0\)

Expanding by grouping the first two terms and last two terms.
\(\left[x^{2}+2 x+x+2\right]\left[x^{2}+4 x+3 x+12\right]+1=0\)
\(\Rightarrow\left[x^{2}+3 x+2\right]\left[x^{2}+7 x+12\right]+1=0\)

We find no similarity, so let us write the given equation considering different groupings as:
\([(x+1)(x+4)][(x+2)(x+3)]+1=0\)
\(\Rightarrow\left[x^{2}+4 x+x+4\right]\left[x^{2}+3 x+2 x+6\right]+1=0\)
\(\Rightarrow\left[x^{2}+5 x+4\right]\left[x^{2}+5 x+6\right]+1=0 \ldots \ldots .(1)\)

Now, we are getting \(x^{2}+5 x\) as common in a product of the group. Let \(x^{2}+5 x=y\), substitute in equation \((1)\), we get:
\((y+4)(y+6)+1=0\)
\(\Rightarrow y^{2}+6 y+4 y+24+1=0\)
\(\Rightarrow y^{2}+10 y+25=0\)

Type VI: Irrational equations reducible to quadratics as:

\(\sqrt{a x+b}=x+k\) or \(\sqrt{a x+b}+\sqrt{c x+d}=k\) or \(\sqrt{a x+b}+\sqrt{c x+d}=\sqrt{e x+f}\).

Method: In this type of equation, we square the given equation on both sides to obtain the equation in quadratic form. The example below explains the method.

Example: \(\sqrt{4-x}+\sqrt{9+x}=5\)

The given equation can be written as:

\(\sqrt{9+x}=5-\sqrt{4-x}\)

Squaring on both sides of the equation, we have:
\(9+x=25+4-x-2 \times 5 \times \sqrt{4-x}\)
\(\Rightarrow 2 x-20=-10 \sqrt{4-x}\)
\(\Rightarrow 2(x-10)=-10 \sqrt{4-x}\)
\(\Rightarrow(x-10)=-5 \sqrt{4-x}\)

Again squaring on both sides, we get,
\(x^{2}+100-20 x=25(4-x)\)
\(\Rightarrow x^{2}+100-20 x=100-25 x\)
\(\Rightarrow x^{2}+5 x=0\)

Reducing Cubic Equation to Quadratic Equation

An equation of third-degree is called a cubic equation. The general form of a cubic function is \(f(x)=a x^{3}+b x^{2}+c x^{1}+d\) where \(a, b\) and \(c\) are the coefficients of the variable and \(d\) is the constant.

We solve a cubic equation by reducing it to a quadratic equation. Then we follow any convenient method to factorise the equation, either by using quadratic formula or splitting the middle term method. The cubic equation will have three roots, like a quadratic equation with two real roots.

Example: \(x^{3}-5 x^{2}-2 x+24=0\)

By trial and error method, put \(x=-2\), we get,

\((-2)^{3}-5(-2)^{2}-2(-2)+24=0\)

\(\Rightarrow-8-20+4+24=0\)

\(\Rightarrow 0=0\)

So one of the roots of the equation is \(-2\)

This states that if \(x=s\) is a solution, then \((x-s)\) is a factor that can be taken out of the equation. For this situation, \(s=-2\), and so \((x+2)\) is a factor we get, \((x+2)\left(x^{2}+a x+b\right)=0\).

Learn Equations Reducible to Linear Form

Solved Problems on Equations Reducible to Quadratic Equations

Here are some situational problems based on reducible to quadratic equations.

Q.1. Solve: \(x^{6}-9 x^{3}+8=0\).
Ans: Given, \(x^{6}-9 x^{3}+8=0\)
Substituting \(x^{3}=y\) in the given the equation we get,
\(y^{2}-9 y+8=0\)
\(\Rightarrow y^{2}-8 y-y+8=0\)
\(\Rightarrow y(y-8)-1(y-8)=0\)
\(\Rightarrow(y-1)(y-8)=0\)
\(\Rightarrow(y-1)=0,(y-8)=0\)
\(\Rightarrow y=1, y=8\)

Now put back the values to get the value of \(x\)
\(y=1 \Rightarrow x^{3}=1^{3} \Rightarrow x=1\)
\(y=8 \Rightarrow x^{3}=2^{3} \Rightarrow x=2\)
Thus, the solution of the equation is \(x=(1,2)\).

Q.2. Solve for \(x: 5 x-\frac{35}{x}=18, x \neq 0\).
Ans: Given, \(5 x-\frac{35}{x}=18\)
The given equation may be written as follows by taking the LCM of the denominators,
\(\frac{5 x^{2}-35}{x}=18\)
\(\Rightarrow 5 x^{2}-35=18 x\)
\(\Rightarrow 5 x^{2}-18 x-35=0\)
Using the splitting of the middle term method, we get:

\(\Rightarrow 5 x^{2}-25 x+7 x-35=0\)
\(\Rightarrow 5 x(x-5)+7(x-5)=0\)
\(\Rightarrow(5 x+7)(x-5)=0\)
\(\Rightarrow(5 x+7)=0,(x-5)=0\)
\(\Rightarrow x=\frac{-7}{5}, x=5\)
Hence, the solution set for the given equation is \(\left(\frac{-7}{5}, 5\right)\)

Q.3. Solve for \(z: 3^{z+2}+3^{-z}=10\).
Ans: Given, \(3^{z+2}+3^{-z}=10\)
By using the exponential rule, the given equation may be written as:
\(3^{z} \times 3^{2}+3^{-z}=10\)
Substitute \(3^{z}=y\) in the equation \((1)\)
\(y \times 3^{2}+y^{-1}=10\)
\(\Rightarrow 9 y+\frac{1}{y}=10\)
\(\Rightarrow 9 y^{2}+1=10 y\)
\(\Rightarrow 9 y^{2}-10 y+1=0\)
\(\Rightarrow 9 y^{2}-9 y-y+1=0\)
\(\Rightarrow 9 y(y-1)-1(y-1)=0\)
\(\Rightarrow(9 y-1)(y-1)=0\)
\(\Rightarrow 9 y-1=0, y-1=0\)
\(\Rightarrow y=\frac{1}{9}, y=1\)

Now back substitute in the assumed value to obtain the value of \(z\)
\(3^{z}=3^{-2}, 3^{z}=3^{0}\)
\(\Rightarrow z=-2, z=0\)
Thus, the value of \(z=(-2,0)\).

Q.4. Solve: \(\sqrt{\left(x^{2}+5 x+1\right)}-1=2 x\).
Ans: Given, \(\sqrt{\left(x^{2}+5 x+1\right)}-1=2 x\)
We can write the given equation as,
\(\sqrt{\left(x^{2}+5 x+1\right)}=2 x+1 \ldots \ldots.(1)\)

Squaring equation \((1)\), we get:
\(x^{2}+5 x+1=4 x^{2}+1+4 x\)
\(\Rightarrow-3 x^{2}+x=0\)
\(\Rightarrow 3 x^{2}-x=0\)
\(\Rightarrow x(3 x-1)=0\)
\(\Rightarrow x=0,(3 x-1)=0\)
\(\Rightarrow x=0, x=\frac{1}{3}\)
Thus, the solution of the equation is \(x=0, x=\frac{1}{3}\).

Q.5. Determine the roots of the cubic equation \(2 x^{3}+3 x^{2}-11 x-6=0\) by reducing it to the factor of a quadratic equation.
Ans: Since \(d=6\), then the possible factors are \(1,2,3\) and \(6\).
Now by applying the factor theorem to check the possible values by trial and error.
\(f(1)=2+3-11-6 \neq 0 f(-1)=-2+3+11-6 \neq 0 f(2)=16+12-22-6=0\)
Hence, \(x=2\) is one of the roots.

We can get the other roots of the equation using the synthetic division method. \(\Rightarrow(x-2)\left(a x^{2}+b x+c\right) \Rightarrow(x-2)\left(2 x^{2}+b x+3\right) \Rightarrow(x-2)\left(2 x^{2}+7 x+3\right)\)

We can obtain the other roots of the equation using the splitting of the middle term of a quadratic equation.
\(\Rightarrow(x-2)\left(2 x^{2}+6 x+x+3\right)\)
\( \Rightarrow \left( {x – 2} \right)\left\{ {2x\left( {x + 3} \right) + 1\left( {x + 3} \right)} \right\}\)
\(\Rightarrow(x-2)(2 x+1)(x+3)\)
Therefore, the solutions are \(x=2, x=-\frac{1}{2}\) and \(x=-3\).

Summary

This article studied the definition and standard form of a quadratic equation. This article is exclusively written for the equations that are not in quadratic form but are reduceable to quadratic equations. We learnt six types of equations.

Some forms are biquadratic equations, reciprocal equations, and irrational equations reducible to quadratics. Also, we learned the definition of the cubic equation and solved the cubic equations by reducing them to quadratic equations.

FAQs on Equations Reducible to Quadratic Equations

The most commonly asked queries on equations to quadratic equations are answered here:

Q.1: How to factorise a quadratic equation?
A: To factorise a quadratic equation, first write the given equation in the standard form of a quadratic equation. Then use the quadratic formula to find the factors of the given equation. Also, the equation can be solved by splitting the middle term method. To find the factors by equating the factors to zero and solve the linear equation.

Q.2: What is the standard form of a quadratic equation?
A: Quadratic equations are equations with at least one squared variable. The quadratic equation in standard form is essential when using the quadratic formula to solve it. The standard form of a quadratic equation is \(a x^{2}+b x+c=0\) where \(a, b\) are the coefficients and \(c\) is the constant.

Q.3: What are the uses of a quadratic equation?
A: Actually, quadratic equations are used in our everyday life. Some real-life applications are calculating the area to determine the products, profit or formulating the speed of a vehicle. We can formulate the quadratic equation for the given data in real-life examples and solve them to find the unknown value.

Q.4: Why do we learn quadratic equations?
A: Quadratic functions are essential and unique part of the school curriculum. The values of these equations can be easily calculated. They are advanced to linear functions and give a significant move away from attachment to straight lines.

Q.5: How do you convert an equation to a quadratic equation?
A: Some of the equations of various types can be reduced to quadratic form. We are converting the equation to a quadratic by making the substitution.
For example, \(y^{4}+10 y^{2}+9\)
If we substitute \(y^{2}\) as \(x\), the equation will be reduced to quadratic form \(x^{2}+10 x+9=0\). A quadratic equation that can be solved by any method and back substitute the value to find the actual roots of the equation.