- Written By Sushma_P
- Last Modified 20-07-2022
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
- \(3 x^{2}-5 x+4=0\)
- \(x^{2}+4 x+3=0\)
- 3x(x+8)=-2
- 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.
FAQs
What is an example of a equation reducible to a quadratic equation? ›
By 'reducible' to quadratic, I meant converting the equation to a quadratic one by making a (clever) substitution. Let me show you how. Consider the original equation again: x4 – 10x2 + 9 = 0. If we substitute x2 as y, the equation will look like y2 – 10y + 9 = 0.
What is quadratic equation definition types and examples? ›In math, we define a quadratic equation as an equation of degree 2, meaning that the highest exponent of this function is 2. The standard form of a quadratic is y = ax^2 + bx + c, where a, b, and c are numbers and a cannot be 0. Examples of quadratic equations include all of these: y = x^2 + 3x + 1.
What is an example of a quadratic equation with answers? ›Examples of quadratic equations are: 6x² + 11x – 35 = 0, 2x² – 4x – 2 = 0, 2x² – 64 = 0, x² – 16 = 0, x² – 7x = 0, 2x² + 8x = 0 etc. From these examples, you can note that, some quadratic equations lack the term “c” and “bx.”
What are the 3 types of solving of quadratic equation? ›There are three basic methods for solving quadratic equations: factoring, using the quadratic formula, and completing the square.
What are reducible equations? ›A reducible equation is an equation that can be made simpler by ”reducing” its terms to an equivalent equation that is easier to solve.
What are the 6 ways to solve a quadratic equation? ›- Quadratic Equations and Inequalities.
- Dilations of Quadratic Graphs. ...
- Vocabulary of Quadratic Polynomials. ...
- Solving Quadratic Equations by Factoring. ...
- Solving Quadratic Equations Using Square Roots. ...
- Solving a Quadratic by Completing the Square. ...
- Deriving the Quadratic Formula.
(x + 2)(x - 3) = 0 [standard form: x² - 1x - 6 = 0] (x + 1)(x + 6) = 0 [standard form: x² + 7x + 6 = 0] (x - 6)(x + 1) = 0 [standard form: x² - 5x - 6 = 0] -3(x - 4)(2x + 3) = 0 [standard form: -6x² + 15x + 36 = 0]
How many types of quadratic equations are there give examples? ›There are three different forms of quadratic equations, and they are: Standard form: The standard form of a quadratic equation is represented by y = a x 2 + b x + c where and are just the numbers.
What is Example 1 of quadratic equation? ›For example, 2x2 + x – 300 = 0 is a quadratic equation. Similarly, 2x2 – 3x + 1 = 0, 4x – 3x2 + 2 = 0 and 1 – x2 + 300 = 0 are also quadratic equations. In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree 2, is a quadratic equation.
How do you write a quadratic equation example? ›A quadratic function is a polynomial function of degree 2. So, y = x^2 is a quadratic equation, as is y = 3x^2 + x + 1. All of these are polynomial functions of degree 2.
What is the definition of a quadratic equation? ›
Definitions: A quadratic equation is a second-order polynomial equation in a single variable x ax2+bx+c=0. with a ≠ 0 . Because it is a second-order polynomial equation, the fundamental theorem of algebra guarantees that it has at least one solution.
What are the 5 different ways to solve a quadratic equation? ›- FACTORING. Set the equation equal to zero. ...
- PRINCIPLE OF SQUARE ROOTS. If the quadratic equation involves a SQUARE and a CONSTANT (no first degree term), position the square on one side and the constant on the other side.
- COMPLETING THE SQUARE. ...
- QUADRATIC FORMULA.
Answer: There are various methods by which you can solve a quadratic equation such as: factorization, completing the square, quadratic formula, and graphing. These are the four general methods by which we can solve a quadratic equation.
What are the 5 ways to solve quadratic equation? ›- Factoring.
- Completing the Square.
- Quadratic Formula.
- Graphing.
In mathematics, reduction refers to the rewriting of an expression into a simpler form. For example, the process of rewriting a fraction into one with the smallest whole-number denominator possible (while keeping the numerator a whole number) is called "reducing a fraction".
What is a reducible function? ›REDUCIBLE indicates that a procedure or entry need not be invoked multiple times if the argument(s) stays unchanged, and that the invocation of the procedure has no side effects. For example, a user-written function that computes a result based on unchanging data should be declared REDUCIBLE.
How do you solve a reduction method? ›The elimination method, also called the reduction method, is a method for algebraically solving a system of equations when the two equations have the form ax+by=c. a x + b y = c .
What are the 4 ways to solve an equation? ›We have 4 ways of solving one-step equations: Adding, Substracting, multiplication and division. If we add the same number to both sides of an equation, both sides will remain equal.
How many types of solutions can a quadratic equation have? ›As we have seen, there can be 0, 1, or 2 solutions to a quadratic equation, depending on whether the expression inside the square root sign, (b2 - 4ac), is positive, negative, or zero. This expression has a special name: the discriminant.
What is the formula for solving a quadratic equation? ›The quadratic formula helps us solve any quadratic equation. First, we bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. Then, we plug these coefficients in the formula: (-b±√(b²-4ac))/(2a) . See examples of using the formula to solve a variety of equations.
What are some quadratic expressions? ›
...
Definition of Quadratic Expressions.
| Expressions | Values of a, b, and c | |
|---|---|---|
| 2. | 4x2 - 9 | a = 4 , b = 0 , c = -9 |
| 3. | x2 + x | a = 1 , b = 1 , c = 0 |
| 4. | 6x - 8 | a = 0 , b = 6 , c = -8 |
It is a polynomial equation of degree two that can be written in the form ax2 + bx + c = 0, where a, b, and c are real numbers and a ≠ 0.
What is an example of a not a quadratic equation? ›(iv)x+5=3⇒0x2+x+2=0 which is not a quadratic equation because the coefficient of x2 (a)=0.
What is the best example of quadratic equation? ›- x 2 + x − 30 = 0.
- 5 t 2 + 4 t + 1 = 0.
- 16 x 2 − 4 = 0.
- 3 x 2 + x = 0.
- 5 x 2 = 25.
The term ax2 is called the quadratic term (hence the name given to the function), the term bx is called the linear term, and the term c is called the constant term.
How do you solve standard form examples? ›The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it's pretty easy to find both intercepts (x and y). This form is also very useful when solving systems of two linear equations.
What is the standard form of a quadratic equation? ›A quadratic function is one of the form f(x) = ax2 + bx + c, where a, b, and c are numbers with a not equal to zero. The graph of a quadratic function is a curve called a parabola. Parabolas may open upward or downward and vary in "width" or "steepness", but they all have the same basic "U" shape.
What is an example of a quadratic equation with roots? ›The following are examples of some quadratic equations: 1) x2+5x+6 = 0 where a=1, b=5 and c=6. For every quadratic equation, there can be one or more than one solution. These are called the roots of the quadratic equation.
Where are quadratic equations used? ›Quadratic equations are used in many real-life situations such as calculating the areas of an enclosed space, the speed of an object, the profit and loss of a product, or curving a piece of equipment for designing.
What is an example of irreducible quadratic? ›As a result they cannot be reduced into factors containing only real numbers, hence the name irreducible. Examples include x2+1 or indeed x2+a for any real number a>0, x2+x+1 (use the quadratic formula to see the roots), and 2x2−x+1.
What are the 5 examples of quadratic expression? ›
- x 2 + x − 30 = 0.
- 5 t 2 + 4 t + 1 = 0.
- 16 x 2 − 4 = 0.
- 3 x 2 + x = 0.
- 5 x 2 = 25.
e.g. x2- 2x - 8 = 0 and 5x2 + 3x - 2 = 0 are affected quadratic equations.
What are the example equations reducible to a pair of linear equations in two variables? ›An example for the pair of linear equations in two variables is 5x =y and -7x+2y+3 =0.
What are irreducible factors examples? ›Factors that cannot be further factorized over another real number are called irreducible factors. For example: Factorization of m2+5 gives m=?(-5), which cannot be factorized further, and the square root of a negative number cannot be found; thus, m2+5 is called an irreducible factor.
What is an example of irreducible? ›Example: 2x + 2 is irreducible in Q[x]. Note that 2x + 2 = 2 (x + 1) where 2 is a unit in Q[x]. Since 2 is a unit in Z[x], 2x + 2 is reducible in Z[x].
What are irreducible elements examples? ›Example: In Z, a = -7 is irreducible because we can only write -7 = (-1)(7) or -7 = (1)(-7) and in both cases one is a unit.
Which expressions are quadratic? ›A quadratic expression, or quadratic equation, is a polynomial equation with a power of two being its highest term. That means the variable has a degree of two, or is squared, in the equation.
Which expressions are quadratic expressions? ›A quadratic expression is an expression that can be written in the form ax² + bx + c, where a, b, and c are numbers and x is a variable; this is known as the standard form.
What are real examples of quadratic functions? ›Throwing a ball, shooting a cannon, diving from a platform and hitting a golf ball are all examples of situations that can be modeled by quadratic functions. In many of these situations you will want to know the highest or lowest point of the parabola, which is known as the vertex.
What are not examples of a quadratic equation? ›- bx − 6 = 0 is NOT a quadratic equation because there is no x2 term.
- x3 − x2 − 5 = 0 is NOT a quadratic equation because there is an x3 term (not allowed in quadratic equations).
What is an example of a linear equation in two variables and write any 2 solutions of that equation? ›
An equation is said to be linear equation in two variables if it is written in the form of ax + by + c=0, where a, b & c are real numbers and the coefficients of x and y, i.e a and b respectively, are not equal to zero. For example, 10x+4y = 3 and -x+5y = 2 are linear equations in two variables.
What is an example 4 pair of linear equations in two variables? ›Some examples of a pair of linear equations in two variables are: 2x + 3y – 7 = 0 and 9x – 2y + 8 = 0. 5x = y and –7x + 2y + 3 = 0.
What are 2 examples of linear equations in one variable? ›The linear equations in one variable is an equation which is expressed in the form of ax+b = 0, where a and b are two integers, and x is a variable and has only one solution. For example, 2x+3=8 is a linear equation having a single variable in it. Therefore, this equation has only one solution, which is x = 5/2.