Solve Quadratic equations 0=4x^3+25x^2-56x Tiger Algebra Solver (2024)

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

0-(4*x^3+25*x^2-56*x)=0

Step by step solution :

Step 1 :

Equation at the end of step 1 :

 0 - (((4 • (x3)) + 52x2) - 56x) = 0 

Step 2 :

Equation at the end of step 2 :

 0 - ((22x3 + 52x2) - 56x) = 0 

Step 3 :

Step 4 :

Pulling out like terms :

4.1 Pull out like factors:

-4x3 - 25x2 + 56x=-x•(4x2 + 25x - 56)

Trying to factor by splitting the middle term

4.2Factoring 4x2 + 25x - 56

The first term is, 4x2 its coefficient is 4.
The middle term is, +25x its coefficient is 25.
The last term, "the constant", is -56

Step-1 : Multiply the coefficient of the first term by the constant 4-56=-224

Step-2 : Find two factors of -224 whose sum equals the coefficient of the middle term, which is 25.

-224+1=-223
-112+2=-110
-56+4=-52
-32+7=-25
-28+8=-20
-16+14=-2
-14+16=2
-8+28=20
-7+32=25That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step2above, -7 and 32
4x2 - 7x+32x - 56

Step-4 : Add up the first 2 terms, pulling out like factors:
x•(4x-7)
Add up the last 2 terms, pulling out common factors:
8•(4x-7)
Step-5:Add up the four terms of step4:
(x+8)•(4x-7)
Which is the desired factorization

Equation at the end of step 4 :

 -x • (4x - 7) • (x + 8) = 0 

Step 5 :

Theory - Roots of a product :

5.1 A product of several terms equals zero.When a product of two or more terms equals zero, then at least one of the terms must be zero.We shall now solve each term = 0 separatelyIn other words, we are going to solve as many equations as there are terms in the productAny solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation:

5.2Solve:-x = 0Multiply both sides of the equation by (-1) : x = 0

Solving a Single Variable Equation:

5.3Solve:4x-7 = 0Add 7 to both sides of the equation:
4x = 7
Divide both sides of the equation by 4:
x = 7/4 = 1.750

Solving a Single Variable Equation:

5.4Solve:x+8 = 0Subtract 8 from both sides of the equation:
x = -8

Supplement : Solving Quadratic Equation Directly

Solving 4x2+25x-56 = 0 directly 

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex:

6.1Find the Vertex ofy = 4x2+25x-56Parabolas have a highest or a lowest point called the Vertex.Our parabola opens up and accordingly has a lowest point (AKA absolute minimum).We know this even before plotting "y" because the coefficient of the first term,4, is positive (greater than zero).Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x-intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.For any parabola,Ax2+Bx+C,the x-coordinate of the vertex is given by -B/(2A). In our case the x coordinate is -3.1250Plugging into the parabola formula -3.1250 for x we can calculate the y-coordinate:
y = 4.0 * -3.12 * -3.12 + 25.0 * -3.12 - 56.0
or y = -95.062

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = 4x2+25x-56
Axis of Symmetry (dashed) {x}={-3.12}
Vertex at {x,y} = {-3.12,-95.06}
x-Intercepts (Roots) :
Root 1 at {x,y} = {-8.00, 0.00}
Root 2 at {x,y} = { 1.75, 0.00}

Solve Quadratic Equation by Completing The Square

6.2Solving4x2+25x-56 = 0 by Completing The Square.Divide both sides of the equation by 4 to have 1 as the coefficient of the first term :
x2+(25/4)x-14 = 0

Add 14 to both side of the equation :
x2+(25/4)x = 14

Now the clever bit: Take the coefficient of x, which is 25/4, divide by two, giving 25/8, and finally square it giving 625/64

Add 625/64 to both sides of the equation :
On the right hand side we have:
14+625/64or, (14/1)+(625/64)
The common denominator of the two fractions is 64Adding (896/64)+(625/64) gives 1521/64
So adding to both sides we finally get:
x2+(25/4)x+(625/64) = 1521/64

Adding 625/64 has completed the left hand side into a perfect square :
x2+(25/4)x+(625/64)=
(x+(25/8))(x+(25/8))=
(x+(25/8))2
Things which are equal to the same thing are also equal to one another. Since
x2+(25/4)x+(625/64) = 1521/64 and
x2+(25/4)x+(625/64) = (x+(25/8))2
then, according to the law of transitivity,
(x+(25/8))2 = 1521/64

We'll refer to this Equation as Eq. #6.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
(x+(25/8))2 is
(x+(25/8))2/2=
(x+(25/8))1=
x+(25/8)

Now, applying the Square Root Principle to Eq.#6.2.1 we get:
x+(25/8)= 1521/64

Subtract 25/8 from both sides to obtain:
x = -25/8 + √ 1521/64

Since a square root has two values, one positive and the other negative
x2 + (25/4)x - 14 = 0
has two solutions:
x = -25/8 + √ 1521/64
or
x = -25/8 - √ 1521/64

Note that 1521/64 can be written as
1521 / √64which is 39 / 8

Solve Quadratic Equation using the Quadratic Formula

6.3Solving4x2+25x-56 = 0 by the Quadratic Formula.According to the Quadratic Formula,x, the solution forAx2+Bx+C= 0 , where A, B and C are numbers, often called coefficients, is given by :

-B± √B2-4AC
x = ————————
2A
In our case,A= 4
B= 25
C=-56
Accordingly,B2-4AC=
625 - (-896) =
1521
Applying the quadratic formula :

-25 ± √ 1521
x=———————
8
Can 1521 be simplified ?

Yes!The prime factorization of 1521is
3•3•13•13
To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).

1521 =√3•3•13•13 =3•13•√ 1 =
±39 •√ 1 =
±39

So now we are looking at:
x=(-25±39)/8

Two real solutions:

x =(-25+√1521)/8=(-25+39)/8= 1.750

or:

x =(-25-√1521)/8=(-25-39)/8= -8.000

Three solutions were found :

  1. x = -8
  2. x = 7/4 = 1.750
  3. x = 0
Solve Quadratic equations 0=4x^3+25x^2-56x Tiger Algebra Solver (2024)

FAQs

Are there 2 answers for the quadratic formula? ›

To determine the number of solutions of each quadratic equation, we will look at its discriminant. Since the discriminant is positive, there are 2 real solutions to the equation. Since the discriminant is negative, there are 2 complex solutions to the equation.

How do you find the answer to 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 is the value of the discriminant of the quadratic equation 2 2x 4x 3 0? ›

The given equation is of the form ax2 + bx + c = 0 where a = 2 b = – 4 andc = 3. Therefore the discriminantb2 – 4ac = – 42 – 4 × 2 × 3 = 16 – 24 = – 8 < 0So the given equation has no real roots.

What is the nature of the roots of the quadratic equation 2x2 4x 3 0? ›

Thus, roots of the equation are imaginary.

Do quadratics have 2 answers? ›

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.

How to solve a quadratic equation without a calculator? ›

First, you find a common factor between a, b and c. Then, you find the 2 numbers that multiplied give you c and added give you b. If a=1, then the additive inverse of those 2 numbers are the answers. If a does not equal 1, you decompose b into the 2 numbers that multiplied give you c and added give you b.

What is the easiest way to find a quadratic equation? ›

Factoring is the first of the three methods of solving quadratic equations. It is often the fastest way to solve a quadratic equation, so usually should be attempted before any other method. This method relies on the fact that if two expressions multiply to zero, then at least one of them must be zero.

How to solve a quadratic equation easily? ›

Set the equation equal to zero. If the quadratic side is factorable, factor, then set each factor equal to zero. 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. Then take the square root of both sides.

How to find the nature of roots? ›

To determine the nature of roots of quadratic equations (in the form ax^2 + bx +c=0) , we need to calculate the discriminant, which is b^2 – 4 a c. When the discriminant is greater than zero, the roots are unequal and real. When the discriminant is equal to zero, the roots are equal and real.

What is the discriminant and how many solutions would this quadratic have? ›

The discriminant is the part of the quadratic formula underneath the square root symbol: b²-4ac. The discriminant tells us whether there are two solutions, one solution, or no solutions.

How can you tell from the discriminant how many real number solutions a quadratic has? ›

It determines the number and the type of solutions that a quadratic equation has. If the discriminant is positive, there are 2 real solutions. If it is 0 , there is 1 real repeated solution. If the discriminant is negative, there are 2 complex solutions (but no real solutions).

What is the formula for the discriminant? ›

The discriminant formula is d=b^2-4ac given the equation of the quadratic is f(x)=ax^2+bx+c. The formula derives from the quadratic formula. Moreover, the discriminant tells how many solutions to the equation there are as well as if the solutions are real or imaginary.

What is the standard form of the quadratic equation? ›

ax2+bx+c = 0 is the standard form of a quadratic equation.

Which of these is a quadratic equation? ›

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.

How many solutions exist for a quadratic? ›

The quadratic equation of the standard form ax2+bx+c has two solutions. The solutions or roots of quadratic equations can be positive/negative, same/distinct, and real or imaginary.

Why is there two quadratic formulas? ›

Quadratics usually have two answers because they can be reduced to a square root, in a suitable sense, and there are typically two square roots of any given value (since x2=y x 2 = y is equivalent to (x−√y)(x+√y)=0 ( x − y ) ( x + y ) = 0 , and there are two choices of which factor on the left to make zero).

How to tell if an equation has two solutions? ›

if the discriminant is positive. , then the quadratic equation has two solutions. if the discriminant is equal. , then the quadratic equation has one solution.

What quadratic equation has two real solutions? ›

For the quadratic equation ax2 + bx + c = 0, the expression b2 – 4ac is called the discriminant. The value of the discriminant shows how many roots f(x) has: - If b2 – 4ac > 0 then the quadratic function has two distinct real roots.

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