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Solving a quadratic equation using grouping

Use grouping to factor and solve the quadratic equation: 4 x 2 + 15 x + 9 = 0.

First, multiply a c : 4 ( 9 ) = 36. Then list the factors of 36.

1 36 2 18 3 12 4 9 6 6

The only pair of factors that sums to 15 is 3 + 12. Rewrite the equation replacing the b term, 15 x , with two terms using 3 and 12 as coefficients of x . Factor the first two terms, and then factor the last two terms.

4 x 2 + 3 x + 12 x + 9 = 0 x ( 4 x + 3 ) + 3 ( 4 x + 3 ) = 0 ( 4 x + 3 ) ( x + 3 ) = 0

Solve using the zero-product property.

( 4 x + 3 ) ( x + 3 ) = 0 ( 4 x + 3 ) = 0 x = 3 4 ( x + 3 ) = 0 x = 3

The solutions are 3 4 , and −3. See [link] .

Coordinate plane with the x-axis ranging from negative 6 to 2 with every other tick mark labeled and the y-axis ranging from negative 6 to 2 with each tick mark numbered. The equation: four x squared plus fifteen x plus nine is graphed with its x-intercepts: (-3/4,0) and (-3,0) plotted as well.
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Solve using factoring by grouping: 12 x 2 + 11 x + 2 = 0.

( 3 x + 2 ) ( 4 x + 1 ) = 0 , x = 2 3 , x = 1 4

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Solving a polynomial of higher degree by factoring

Solve the equation by factoring: −3 x 3 5 x 2 2 x = 0.

This equation does not look like a quadratic, as the highest power is 3, not 2. Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out x from all of the terms and then proceed with grouping.

−3 x 3 5 x 2 2 x = 0 x ( 3 x 2 + 5 x + 2 ) = 0

Use grouping on the expression in parentheses.

x ( 3 x 2 + 3 x + 2 x + 2 ) = 0 x [ 3 x ( x + 1 ) + 2 ( x + 1 ) ] = 0 x ( 3 x + 2 ) ( x + 1 ) = 0

Now, we use the zero-product property. Notice that we have three factors.

x = 0 x = 0 3 x + 2 = 0 x = 2 3 x + 1 = 0 x = −1

The solutions are 0 , 2 3 , and −1.

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Solve by factoring: x 3 + 11 x 2 + 10 x = 0.

x = 0 , x = −10 , x = −1

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Using the square root property

When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property    , in which we isolate the x 2 term and take the square root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the x 2 term so that the square root property can be used.

The square root property

With the x 2 term isolated, the square root property states that:

if x 2 = k , then x = ± k

where k is a nonzero real number.

Given a quadratic equation with an x 2 term but no x term, use the square root property to solve it.

  1. Isolate the x 2 term on one side of the equal sign.
  2. Take the square root of both sides of the equation, putting a ± sign before the expression on the side opposite the squared term.
  3. Simplify the numbers on the side with the ± sign.

Solving a simple quadratic equation using the square root property

Solve the quadratic using the square root property: x 2 = 8.

Take the square root of both sides, and then simplify the radical. Remember to use a ± sign before the radical symbol.

x 2 = 8 x = ± 8 = ± 2 2

The solutions are 2 2 , −2 2 .

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Solving a quadratic equation using the square root property

Solve the quadratic equation: 4 x 2 + 1 = 7.

First, isolate the x 2 term. Then take the square root of both sides.

4 x 2 + 1 = 7 4 x 2 = 6 x 2 = 6 4 x = ± 6 2

The solutions are 6 2 , and 6 2 .

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Solve the quadratic equation using the square root property: 3 ( x 4 ) 2 = 15.

x = 4 ± 5

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Completing the square

Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation    known as completing the square    . Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a , must equal 1. If it does not, then divide the entire equation by a . Then, we can use the following procedures to solve a quadratic equation by completing the square.

Questions & Answers

can you solve it step b step
Ching Reply
what is linear equation with one unknown 2x+5=3
Joan Reply
-4
Joel
x=-4
Joel
x=-1
Joan
I was wrong. I didn't move all constants to the right of the equation.
Joel
x=-1
Cristian
Adityasuman x= - 1
Aditya
what is the VA Ha D R X int Y int of f(x) =x²+4x+4/x+2 f(x) =x³-1/x-1
Shadow Reply
can I get help with this?
Wayne
Are they two separate problems or are the two functions a system?
Wilson
Also, is the first x squared in "x+4x+4"
Wilson
x^2+4x+4?
Wilson
thank you
Wilson
Please see ***imgur.com/a/lpTpDZk for solutions
Wilson
f(x)=x square-root 2 +2x+1 how to solve this value
Marjun Reply
factor or use quadratic formula
Wilson
what is algebra
Ige Reply
The product of two is 32. Find a function that represents the sum of their squares.
Paul
if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta
Martin Reply
how to compute this 1. g(1-x) 2. f(x-2) 3. g (-x-/5) 4. f (x)- g (x)
Yanah Reply
hi
John
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Grace
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John
not much For functions, there are two conditions for a function to be the inverse function:   1--- g(f(x)) = x for all x in the domain of f     2---f(g(x)) = x for all x in the domain of g Notice in both cases you will get back to the  element that you started with, namely, x.
Grace
sin theta=3/4.prove that sec square theta barabar 1 + tan square theta by cosec square theta minus cos square theta
Umesh Reply
acha se dhek ke bata sin theta ke value
Ajay
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Ajay
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Yh
Idowu
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Nyemba
functions
Siyabonga
trigonometry
Ganapathi
differentiation doubhts
Ganapathi
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Ganapathi
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Brittany
Prove that 4sin50-3tan 50=1
Sudip Reply
False statement so you cannot prove it
Wilson
f(x)= 1 x    f(x)=1x  is shifted down 4 units and to the right 3 units.
Sebit Reply
f (x) = −3x + 5 and g (x) = x − 5 /−3
Sebit
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Marty Reply
I want to know partial fraction Decomposition.
Adama Reply
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Yazidu Reply
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Sumanth Reply
Practice Key Terms 7

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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