This year 11 specialist Victorian Math question uses algebra to solve quadratic equations?

Question

Jan-Magnus Økland · Answer

[math]ax^2+bx+c=a((x +\frac{b}{2a})^2-(\frac{d}{2a})^2)[/math], where [math]d^2=b^2-4ac[/math]. The identity to use is [math](A+B)(A-B)=A^2-B^2[/math] in reverse: [math](x +\frac{b}{2a})^2-(\frac{d}{2a})^2=(x +\frac{b}{2a}-\frac{d}{2a})(x +\frac{b}{2a}+\frac{d}{2a})[/math]. Factoring and solving are equivalent since [math]x-r=0[/math] has solution [math]x=r[/math], and [math](x-r)(x-s)=0[/math] has solutions [math]x=r,s[/math]

Anonymous · Answer

To understand the different methods, let us take an example.
Solving: [math]x^2 - 7x + 12 = 0[/math]

Factoring
[math]x^2 - 7x + 12 = 0[/math]

[Splitting the middle term] 
[math]x^2 - 3x - 4x + 12 = 0[/math]

[Taking terms pairwise]
[math]x(x-3) -4(x-3) = 0[/math]

[Pulling out the common term]
[math](x-3)(x-4) = 0[/math]
Or,
[math]x = 3 [/math]or [math]x = 4[/math] are roots of the equation.

Completing the square
The general principle requires the generic quadratic equation [math]ax^2 + bx + c = 0[/math] be re-written in the form a complete square.
[math]x^2 - 7x + 12 = 0[/math] 
can be re-written as
[math]x^2 - 2(x)(7/2) + (7/2)^2 + 12 = (7/2)^2 [/math]

[math](x - 7/2)^2 = 49/4 - 12 [/math]

[math]\implies (x - 7/2)^2 = 1/4[/math]

The roots are [math]x = +1/2 + 7/2 = 4 [/math] or [math]-1/2 + 7/2 = 3[/math]

Quadratic formula
The general formula is given by [math]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/math]

For [math]x^2 - 7x + 12 = 0,
 x = \frac{+7\pm\sqrt{7^2-4(1)(12)}}{2} [/math]

[math]\implies x = \frac{7 \pm \sqrt{1}}{2} [/math]

[math]\implies x = \frac{7 \pm 1}{2}[/math]

The roots are [math]x = 4[/math] or [math]x = 3[/math]

Graphing the Equations
Graphing a quadratic equation results in a parabola, with the zeroes indicated by the intersection of the curve with the x-axis.

The graph of [math]x^2 - 7x + 12[/math] intersects x-axis at 3 & 4 respectively.

Marjan Tanjin · Answer

Solving quadratic equations by hand is an important skill for a few reasons:

1. Building mathematical problem-solving skills: Learning to solve quadratic equations by hand can help build mathematical problem-solving skills. This involves breaking down a complex problem into smaller steps, identifying patterns, and finding a solution. These skills are useful not just in math, but in other areas of life as well.
2. Understanding mathematical concepts: Solving quadratic equations by hand helps to understand the underlying mathematical concepts, such as factoring, completing the square, and the quadratic formula. This understanding can help build a stronger foundation for more advanced math concepts and problem-solving.
3. Practical applications: Quadratic equations have many practical applications in fields such as physics, engineering, finance, and computer science. Being able to solve these equations by hand allows one to understand and solve problems in these fields, without relying on a calculator or computer program.
4. Building confidence: Finally, solving quadratic equations by hand can help build confidence in one's math skills. It can be satisfying to solve a complex problem on your own, and this can help build motivation and interest in math.
Overall, while there are many tools available for solving quadratic equations (such as calculators and computer programs), learning to solve them by hand is an important skill that can help build problem-solving skills, deepen understanding of mathematical concepts, and have practical applications in various fields.

Nikola Pavkovic · Answer

Well, let’s look at a quadratic very quickly in a graphical representation:

This is [math]y = x²+2x-3[/math]

Let’s consider [math]x²+2x-3 = 0[/math] very quickly.
Factorize and we get: [math](x+3)(x-1) = 0
[/math]Two solutions: [math]x = -3, x = 1[/math]

But look at yonder parabola and ye will see:

Our solutions correspond to the intercepts on the x-axis. Because, consider what we just did, we solved for “which x coordinates of this equation have a y position at zero?”. And for parabolas, that’s often two different x coordinates. However, parabolas have a vertex:

This is where you will find only one solution, if you solved for:

[math]x²+2x-3 = -4[/math]

You would find one answer: -1.

As a sidenote:

If you tried to solve for [math]x²+2x-3 = -5[/math]

You would delve into the mysterious and confusing world of “imaginary” and complex numbers, the result of square root of a negative number, and a topic I’m not familiar with.

Hope this helps!

Ginger Eaves · Answer

Normally these concepts are taught in a first year algebra class in high school, but in my 45 years of teaching, with the last 23 teaching college, most of my students were unable to do 4th grade arithmetic, and the college had 9 hours of remedial math prior to college algebra. My students could not do the work that my 8th grade Algebra I students could do in the 90s. It is sad!

Mohammad Afzaal Butt · Answer

[math]	ext{Let the quadratic equation is }[/math]

[math]a x^2 + b x + c = 0[/math]

[math]\implies x^2 + \left(\dfrac{b}{a}ight) x + \dfrac{c}{a} = 0[/math]

[math]\implies x^2 + 2\left(\dfrac{b}{2 a}ight)x + \dfrac{b^2}{4 a^2}  - \dfrac{b^2}{4 a^2} + \dfrac{c}{a} = 0[/math]

[math]\implies \left( x + \dfrac{b}{2 a}ight)^2 - \dfrac{b^2 - 4 a c}{4 a^2} = 0[/math]

[math]\implies \left( x + \dfrac{b}{2 a}ight)^2 -\left( \dfrac{\sqrt{b^2 - 4 a c}}{2 a}ight)^2 = 0[/math]

[math] \implies \left( x + \dfrac{b}{2 a}-\dfrac{\sqrt{b^2 - 4 a c}}{2 a}ight)\left( x + \dfrac{b}{2 a} +  \dfrac{\sqrt{b^2 - 4 a c}}{2 a}ight) = 0[/math]

[math]\implies \left( x + \dfrac{b}{2 a}\pm\dfrac{\sqrt{b^2 - 4 a c}}{2 a}ight)= 0[/math]

[math]\implies x = \dfrac{- b\pm\sqrt{b^2 - 4 a c}}{2 a}[/math]

Robert Paxson · Answer

The general form for the quadratic equation is:

[math]ax^2+bx+c=0[/math]

To solve this equation algebraically:

[math]a(x^2+\frac{b}{a}x)=-c[/math]

[math]a(x^2+\frac{b}{a}x+[\frac{b}{2a}]^2-[\frac{b}{2a}]^2)=-c[/math]

[math]a(x^2+\frac{b}{a}x+[\frac{b}{2a}]^2)-a[\frac{b}{2a}]^2=-c[/math]

[math]a(x+\frac{b}{2a})^2=a[\frac{b}{2a}]^2-c[/math]

[math](x+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{c}{a}[/math]

[math](x+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}[/math]

[math](x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}[/math]

[math]x+\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}[/math]

[math]x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}[/math]

[math]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/math]

The equation for [math]x[/math] that we found using algebra is called the quadratic formula. The algebra that we used above on the parameters [math]a[/math], [math]b[/math], and [math]c[/math], can be applied to any quadratic equation with specific numerical values for these parameters, in the way shown above, to find the required values of [math]x[/math] that solve that particular equation.

For example, suppose we need to solve:

[math]8x^2+22x+15=0[/math]

Then, using algebra, which follows the method given above, we get:

[math]8x^2+22x+15=0[/math]

[math]8(x^2+\frac{11}{4}x)=-15[/math]

[math]8(x^2+\frac{11}{4}x+[\frac{11}{8}]^2-[\frac{11}{8}]^2)=-15[/math]

[math]8(x^2+\frac{11}{4}x+[\frac{11}{8}]^2)-8[\frac{11}{8}]^2=-15[/math]

[math]8(x+\frac{11}{8})^2=\frac{121}{8}-\frac{(15)(8)}{8}[/math]

[math]8(x+\frac{11}{8})^2=\frac{121-120}{8}[/math]

[math](x+\frac{11}{8})^2=\frac{1}{64}[/math]

[math]x+\frac{11}{8}=\pm\sqrt{\frac{1}{64}}[/math]

[math]x+\frac{11}{8}=\pm\frac{1}{8}[/math]

[math]x=-\frac{11}{8}\pm\frac{1}{8}[/math]

[math]x=-\frac{12}{8}[/math], [math]x=-\frac{10}{8}[/math]

[math]x=-\frac{3}{2}[/math], [math]x=-\frac{5}{4}[/math]

Using algebra is a lot of work!

A much easier method is factoring:

[math]8x^2+22x+15=0[/math]

[math](2x+3)(4x+5)=0[/math]

[math]2x+3=0[/math] or [math]4x+5=0[/math]

[math]x=-\frac{3}{2}[/math], [math]x=-\frac{5}{4}[/math]

or, using the quadratic formula that we derived above:

[math]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/math]

[math]x=\frac{-22\pm\sqrt{22^2-4(8)(15)}}{16}[/math]

[math]x=\frac{-22\pm\sqrt{4}}{16}[/math]

[math]x=\frac{-11\pm\,1}{8}[/math]

[math]x=\frac{-12}{8}[/math], [math]x=\frac{-10}{8}[/math]

[math]x=-\frac{3}{2}[/math], [math]x=-\frac{5}{4}[/math]

Paul Holloway · Answer

Real solutions to a quadratic equation are solutions which are real numbers.

Real numbers:

Real numbers are numbers that can represent real things. We learn about them in school, numbers like 1, 2, 3… all the way up to infinity. Also 0 and -1, -2, -3… all the way down to negative infinity are real numbers. Fractions are also real numbers as are irrational numbers. So 1/2, 1/5, 1/10 along with pi or e or the square root of 2 are all real numbers.

Here is a representation of the real numbers on a number line:

Because a negative number multiplied by itself is a positive number there is no real number that represents the square root of a negative number. That is to say there is no real value for x such that x^2 = -1. You can’t point to a position on the number line and say here is x.

Imaginary numbers:

But what if there was a number that when multiplied by itself came to a negative number? Let’s imagine such numbers exist and let’s say the square root of -1 is i where i stands for imaginary. Does mathematics allow for this or will mathematics break down?

i + i = 2i well so far we can add imaginary numbers.

2i - i = i, we can subtract them well enough.

2i/2 = i, we can divide them well enough.

i * i = -1, because of our definition of i.

So far, so good. It turns out we can do the same sort of operations on imaginary numbers as we can on real numbers.

We can draw a similar number line for imaginary numbers but let’s put this number line at right angles to the real number line.

Complex Numbers:

We know we can add or subtract real numbers and we just learned we can add or subtract imaginary numbers but can we add real numbers to imaginary numbers or subtract them? It turns out we can and the maths works just as well. We call these combinations of real and imaginary numbers complex numbers.

2 + 3i is a complex number. 2 - 3i is also a complex number. -2 + 3i is likewise a complex number.

What happens if we try to square a complex number though?

(2+3i)^2 = (2+3i)(2+3i) = 2^2 + (3i)^2 + 2(2)(3i) = 4 + 9i^2 + 12i but we know i^2 = -1 so 9i^2 becomes 9(-1) which is -9. So (2+3i)^2 becomes 4 - 9 + 12i which is -5 + 12i which is also a complex number. If we think of the real number line and the imaginary number line as the axes on a cartesian graph we can plot the position of a complex number on the complex plane.

Thus 1+i would be represented thus:

Now it turns out that we can plot any real or imaginary number on this graph, that is to say, the complex plane. This means all real and imaginary numbers can be considered a subset of complex numbers. If it helps, we can think of i as a complex number (0 + 1i) or we can think of 1 as a complex number (1 + 0i).

Now that we have an idea of what real, imaginary and complex numbers are let’s look again at your question.

The real solutions to a quadratic equation are solutions to that equation that are real.

The solution to x^2 = 1 are 1 and -1. Both these numbers can be plotted on the real number line and so are real numbers and thus are real solutions to the quadratic equation x^2 = 1.

The solution to x^2 = -1 are i and -i. Both these numbers can be plotted on the imaginary number line but not on the real number line so they are not real solutions to the quadratic equation x^2 = -1. Remember a square root of a negative number can’t be plotted anywhere on the real number line.

Now let’s look at the quadratic formula (-b +/- sqrt(b^2–4ac))/2a. For any quadratic equation in the form ax^2 + bx + c = 0.

Note the square root bit. If what is inside the square root is negative we can’t plot the solution on the real number line.

This means that if b^2 %3C 4ac then b^2 - 4ac will be less than zero and therefore negative and its square root will be imaginary. If b^2 = 4ac then b^2 - 4ac = 0 and the square root of zero is zero which means the only real solution is -b/2a.

If b^2 %3E 4ac then b^2 - 4ac will be a positive number and so its square root will be real and there will be two solutions to the quadratic equation.

Thus by looking at the bit under the square root of the quadratic equation (which we call the discriminant) we can figure out whether there are no real solutions, one real solution or two real solutions to any quadratic equation in the form ax^2 + bx + c = 0.

Krishna Raja Sagar · Answer

Quadratic equations are algebraic formulas of the second degree using the formula ax(Square) + bx + c = 0. The term "quadratic" comes from the word "quad," which meaning "square." In other terms, a quadratic equation is a "degree 2 equation." A quadratic equation is utilized in a variety of situations. Did you know that a rocket's course is specified by a quadratic equation when it is launched? A quadratic equation also has many applications in physics, engineering, and astronomy. Quadratic equations are second-degree equations in x with two solutions. The two solutions for x are also known as the quadratic equations' roots and are denoted as (α, β). In the next sections, we will study more about the roots of a quadratic equation.

What Exactly Is a Quadratic Equation?

An algebraic statement of the second degree in x is a quadratic equation (https://www.doubtnut.com/learn/english/class-10/maths/chapter/quadriatic_equations). A quadratic equation has the usual form ax(Square) + bx + c = 0, where a and b are the coefficients, x is the variable, and c is the constant term. The first requirement for an equation to be a quadratic equation is that the coefficient of x(Square) is not zero (a ≠ 0). When formulating a quadratic equation in standard form, the x(Square) term comes first, then the x term, and lastly the constant term. The numeric values a, b, and c are typically represented as integral values rather than fractions or decimals.

Formula for a Quadratic Equation

The Quadratic Formula is the most basic method for determining the roots of a quadratic problem. Certain quadratic equations cannot be easily factorized, and in these cases, we may utilize this quadratic formula to determine the roots as quickly as feasible. The roots of the quadratic equation also aid in determining the sum and product of the roots of the quadratic equation. The quadratic formula's two roots are provided as a single statement. The positive and negative signs can be employed alternately to produce the equation's two different roots.

Important Quadratic Equation Solving Formulas

The collection of essential formulae below can be used to solve quadratic problems. A quadratic equation has the usual form ax(Square) + bx + c = 0. D = b(Square) - 4ac is the discriminant of the quadratic equation. The roots are genuine and distinct when D %3E 0. The roots are genuine and equivalent when D = 0. The roots for D %3C 0 do not exist, or they are fictitious. x = −b±√b(Square)−4ac/ 2a is the formula for finding the roots of a quadratic equation. The sum of a quadratic equation's roots is α + β = -b/a = - Coefficient of x/Coefficient of x(Square). The product of the quadratic equation's Root is = c/a = Constant term/Coefficient of x(Square). The quadratic equation with roots, is x(Square) - (α + β)x + αβ = 0. The quadratic equation f(x) = ax(Square) + bx + c has a minimum value at x = -b/2a for positive values of a (a %3E 0). The quadratic equation f(x) = ax(Square) + bx + c has a maximum value at x = -b/2a for negative values of a (a %3C 0).

Quadratic Equation Roots

The roots of a quadratic equation are the two x values produced by solving the quadratic equation. The symbols alpha (), and beta () are used to represent the roots of a quadratic equation. These quadratic equation roots are also known as the equation zeros. In this section, we will learn how to determine the nature of the roots of a quadratic equation without actually solving the problem. Also, look at the formulae for calculating the sum and product of the equation's roots.

The Characteristics of Quadratic Equation Roots

The nature of the roots of a quadratic equation can be discovered without actually discovering the equation's roots (α, β). This is accomplished by using the discriminant value, which is included in the formula for solving the quadratic equation. The number b(Square) - 4ac is known as the discriminant of a quadratic equation and is denoted by the letter 'D.' The nature of the quadratic equation's roots can be anticipated using the discriminant value.

Check One Example:

Question: Value(s) of k for which the quadratic equation 2x^(2)-k x +k=0 has equal roots is/are

Answer: https://www.doubtnut.com/question-answer/values-of-k-for-which-the-quadratic-equation-2x2-kx-k0-has-equal-roots-is-are-26861818

The Relationship Between Quadratic Equation Coefficients and Roots

The x(Square) coefficient, x term, and constant term of the quadratic equation ax(Square) + bx + c = 0 are important for learning more about the characteristics of quadratic equation roots. The sum and product of quadratic equation roots may be calculated straight from the equation without having to discover the roots of the quadratic equation. The sum of the quadratic equation's roots equals the inverse of the coefficients of x divided by the coefficient of x(Square). The constant term divided by the coefficient of x(Square) equals the product of the root of the equation. The sum and product of the roots of a quadratic equation ax(Square) + bx + c = 0 are as follows. Sum of the Roots: α + β = -b/a = - Coefficient of/ x2 coefficient. Roots Product: = αβ = c/a = Constant term/Coefficient of x(Square). For the above equation roots, the quadratic equation may also be constructed. If α, β, are the quadratic equation's roots, then the quadratic equation is as follows.

You can learn more about interesting topics such as quadratic equations at doubtnut.com. It would be possible for you to get lots of learning resources that would help in proving to be of much use. You can also get to have a look at the video tutorials that would help in clearing all your queries in the right manner. Therefore, you should try to make sure of downloading the Doubtnut app without any second thought to it.

Musa Ibraahim Sim · Answer

1. Get some basics right before attempting.
2. Don't be afraid to make mistakes.
3. Get an expert to guide your working.
4. Keep on trying and seek help when you are stuck.
5. Believe in yourself that you can do it. It may not be now but eventually you can master it.
6. Write neatly.
7. Be careful of sign change.
8. Practice many questions.
9. Have love for Mathematics.
10. Never give up!

William Jorgensen · Answer

Factoring Methods For Quadratic Equations

1. Factor directly if the quadratic doesn’t have any imaginary or radicals in the roots, this is fastest and easiest. Also, if it is a difference of squares do that first, that is zero work!

[math]x^{2}-5x-14\  \text{Make a chart of -14 (c term) factors:}\ \begin{vmatrix} 1 & -14\ 2 & -7\ -2 & 7\ -1 & 14\ \end{vmatrix}\ \text{Do any of these factors sum to the value of the x coefficient?}\ \text{Yes: }\ 2+(-7)=-5\ \text{Those are your factors:}\ (x+2)(x-7)[/math]

If the [math]x^{2}[/math] term’s coefficient isn’t [math]1[/math] you can still factor, but just using the quadratic equation may be less work, here is how that is done:

[math]2x^{2}+3x−5\ \text{When the }x^{2}\text{ coefficient is not 1:}\ \text{first we go: }x^{2}\text{ coefficient times constant: }2\times-5=-10\ \text{ then we make our chart again:}\ \begin{vmatrix} 1 & -10\ 2 & -5\ -1 & 10\ -2 & 5\ \end{vmatrix}\  \ \text{Do any of these factors sum to the value of the x coefficient?}\  \text{Yes: }\ -2+5=3\ \text{In this case those aren’t the factors, we replace the middle x term:}\ 2x^{2}+3x−5\ 2x^{2}-2x+5x−5\ 2x(x-1)+5(x-1) \text{Then factor the whole }(x-1) \text{ out:}\ (x-1)(2x+5)
[/math]
[math]\textsf{The difference of two squares formula is:}\ a^{2}-b^{2} = (a + b)(a - b)\~\x^{2}-9\ (x+3)(x-3)[/math]
2. If you can solve directly extracting the root, do that.

[math]x^{2}-7\  x^{2}=7\ \sqrt{x^{2}}=\pm\sqrt{7}\ x=\pm\sqrt{7}[/math]
3. We can “complete the square”, this is pretty easy when the polynomial has a coefficient of the [math]x^{2}[/math] term equaling [math]1[/math] and really easy if the coefficient of the [math]x[/math] term is an even number.

[math]\text{Standard form of a quadratic is }ax^{2}+bx+c=0\
\text{To make a perfect square we go: }\
ax^{2}+bx+\left(\frac{b}{2}\right)^2=-c+\left(\frac{b}{2}\right)^2\
\left(x +\frac{b}{2}\right)^2=-c+\left(\frac{b}{2}\right)^2\[/math]
4. Finally, use the quadratic formula (which is actually completing the square) usually it is much easier and less prone to mistakes than directly completing the square.

[math]x = \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\textsf{ when }ax^{2}+bx+c[/math]
5. Also, keep in mind if the coefficient of the [math]x^{2}[/math] term is [math]1[/math], you can just delete all the a terms and use this simpler quadratic equation:

[math]x = \frac{-b\pm \sqrt{b^{2}-4c}}{2}\textsf{ when }1x^{2}+bx+c [/math]

Edward M Young · Answer

So let Q(x) = ax^2 + bx + c = 0

let R(x) = dx^2 + ex + f = 0 = Q(x) for simultaneous equality

Example: Q = x^2 - 5x + 6 = 0 and R = 3x^2 + 7 = 0

For Q= 0, x = 2 and x = 3

For R = 0, x = (i)(sqrt 21)/3 and -(i)(sqrt 21)/3

What does it mean for Q and R to be "simultaneous quadratic equations"? I am

not sure, but use elimination method to remove x^2:

Take R - 3Q to eliminate x^2 terms. Remember that Q = R = 0:

R - 3Q = 0 - 0 = 0 = 3x^2 + 7 - 3x^2 + 15x -18 = 0

= 15x - 11 = 0; 15x = 11; x = 11/15

Plug x = 11/15 into Q and into R to verify concurrence:

for Q(x = 11/15) = (121/225) -11/3 + 6 = 7/3 + (121/225) = (525 + 121)/225= 646/225 = (2)(17)(19)/(3)(3)(5)5) = R/3 = 0

for R: 3(121/225) + 7 = (121/ 75) + (525/75)= 646/75 = (2)(17)(29)/(3)(5)(5) = 3Q = 0

Marco Biagini · Answer

The formula used in solving quadratic equations is known as the Quadratic Formula. A quadratic equation is any equation that can be written in the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable.

The Quadratic Formula is:

x = [-b ± sqrt(b^2 - 4ac)] / (2a)

This formula provides the solutions for x in the quadratic equation. The term under the square root, b^2 - 4ac, is called the discriminant. It tells us whether the roots are real or complex:

1. There are two distinct real roots if the discriminant is positive.

2. If the discriminant is zero, there is exactly one real root (also called a repeated root).

3. If the discriminant is negative, there are two complex roots.

This formula allows you to solve any quadratic equation, provided you know the values of a, b, and c.

Steve Sparling · Answer

Are there three steps?

1. get all the terms on one side and zero on the other
2. 
1. if the leading term is negative multiply both sides by -1
2. if there are fractions multiply both sides by an appropriate number to get rid of the fractions
3. if any of the coefficients are irrational numbers or complex numbers then go straight to 2(c)

3. factor (if you can)
4. 
1. if you cannot factor try completing the square
2. if you don’t like completing the square use the quadratic formula
3. if you don’t like any of the above then download graphing software and do it that way

5. set each linear factor equal to zero and solve
You are right. There are only 3 steps. I somehow feel I missed something. What if it is something like x^2=9. I didn’t have a step for that!

Shreyas Devesh · Answer

For a general quadratic equation say,

ax² + bx + c = 0

It is if and only if the discriminant of the above quadratic equation is greater than or equal to zero that a real solution exists.

For finding possible solutions to a quadratic equation we know of a general formula or the famous quadratic equation -

Here the only way a root can become complex is if the Term in the square root becomes negative (This term is called the discriminant).

Mathematically we say that for real roots b² - 4ac ≥ 0

Neal Schermerhorn · Answer

I didn’t mention one.

But let’s say I mentioned ax^2 + bx + c = 0.

Let’s complete the square. First divide everything by a.

x^2 + (b/a)x + c/a = 0

Move the constant to the other side, and add the square of half of b/a to both sides. (That’s b^2/(4a^2).)

x^2 + (b/a)x + b^2/(4a^2)) = b^2/(4a^2)) - c/a

Left side is a perfect square. And let’s make the right side have a common denominator too.

(x + b/(2a))^2 = (b^2 - 4ac)/(4a^2)

Square root both sides. Don’t forget the plus-minus!

x + b/(2a) = +/- sqrt(b^2 - 4ac))/2a

Subtract b/(2a) from both sides and combine the fractions.

x = (-b +/- sqrt(b^2 - 4ac))/(2a)

That’s how I’d solve it.

Anonymous · Answer

Two quadratic equations will have four roots and since there are only three unknowns in your problem that means one of the four must be already known. Because we are given no information about which one of the four is known there is no alternative but to solve both equations in the usual manner.

David Webb · Answer

By definition an algebraic number is the solution to some polynomial equation with integer (or equivalently rational coefficients ) and also by that same definition all solutions to such polynomials are algebraic numbers. However solutions to non-polynomial equations or polynomial equations with irrational coefficients need not be algebraic numbers. There are an uncountably infinite number of these non algebraic ( transcendental ) numbers including numbers such as pi and e.

Kevin Plymale · Answer

As written?

X-4x+4=0

-%3E -3x+4 = 0

-%3E 4 = 3x

-%3E x=4/3

Now, I PRESUME that you meant this to be a quadratic equation of the form ax^2+bx+c. If so, then it seems you meant:

X^2-4x+4 = 0.

Looking at this, I can tell that (x-2) is a factor (in fact the equation is just (x-2)^2), but let's use the quadratic formula.

Gives us roots of (-b+ sqrt(b^2-4ac))/(2a) and (-b- sqrt(b^2-4ac)/(2a)

With a = 1, b = -4, c = 4.

Our discriminant, b^2-4ac, is then (-4)^2–4(1)(4) = 16–16 = 0.

Since the discriminant is 0, the function has one real root which acts as a double root, since whether we use the first or second root for.ula above, since the discriminant is 0, they are equal.

Thus our double root is (4–0)/2 = 2.

X=2 is our double root, which as notes above means that the equation is just (x-2)^2

Anonymous · Answer

Probably you wanted to ask in how many days will the chapter 'Quadratic Equations' get completed.

For class 11 exams the chapter can be understood with depth in just 12–15 days with a practise of 1.5 hr a day (apart from lecture timings).

For JEE ADVANCE level 1.5 hr practice for almost 21–25 days (excluding lectures) is needed to master it as there are many topics taught extra for the JEE ADVANCE.

And Quadratic Equation is also the most important topic of Mathematics. Without learning this chapter it is really difficult to study Complex numbers or to understand the behaviour of the graphs of various Quadratic functions.