Use quadratic equation to solve
4x/x+4-3/x+3=2/x+2-1/x+1
Answers
Answer:
Solving quadratics by factoring
CCSS Math: HSA.REI.B.4, HSA.REI.B.4b, HSA.SSE.B.3, HSA.SSE.B.3a, HSF.IF.C.8, HSF.IF.C.8a
Learn how to solve quadratic equations like (x-1)(x+3)=0 and how to use factorization to solve other forms of equations.
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What you should be familiar with before taking this lesson
Factoring using the Sum-Product pattern
Factoring by grouping
Factoring special products
What you will learn in this lesson
So far you have solved linear equations, which include constant terms—plain numbers—and terms with the variable raised to the first power, x^1=xx
1
=xx, start superscript, 1, end superscript, equals, x.
You may have also solved some quadratic equations, which include the variable raised to the second power, by taking the square root from both sides.
In this lesson, you will learn a new way to solve quadratic equations. Specifically you will learn
how to solve factored equations like (x-1)(x+3)=0(x−1)(x+3)=0left parenthesis, x, minus, 1, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, equals, 0 and
how to use factorization methods in order to bring other equations ((left parenthesislike x^2-3x-10=0)x
2
−3x−10=0)x, start superscript, 2, end superscript, minus, 3, x, minus, 10, equals, 0, right parenthesis to a factored form and solve them.
Solving factored quadratic equations
Suppose we are asked to solve the quadratic equation (x-1)(x+3)=0(x−1)(x+3)=0left parenthesis, x, minus, 1, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, equals, 0. [Why is this a quadratic equation?]
x^2+2x-3=0
x, start superscript, 2, end superscript, plus, 2, x, minus, 3, equals, 0
This is a product of two expressions that is equal to zero. Note that any xxx value that makes either (x-1)(x−1)left parenthesis, x, minus, 1, right parenthesis or (x+3)(x+3)left parenthesis, x, plus, 3, right parenthesis zero, will make their product zero.
\begin{aligned} (x-1)&(x+3)=0 \\\\ \swarrow\quad&\quad\searrow \\\\ x-1=0\quad&\quad x+3=0 \\\\ x=1\quad&\quad x=-3 \end{aligned}
(x−1)
↙
x−1=0
x=1
(x+3)=0
↘
x+3=0
x=−3
Substituting either x=1x=1x, equals, 1 or x=-3x=−3x, equals, minus, 3 into the equation will result in the true statement 0=00=00, equals, 0, so they are both solutions to the equation.
Now solve a few similar equations on your own.
Solve (x+5)(x+7)=0(x+5)(x+7)=0left parenthesis, x, plus, 5, right parenthesis, left parenthesis, x, plus, 7, right parenthesis, equals, 0.
Choose 1 answer:
Choose 1 answer:
(Choice A)
A
x=5x=5x, equals, 5 and x=7x=7x, equals, 7
(Choice B)
B
x=5x=5x, equals, 5 and x=-7x=−7x, equals, minus, 7
(Choice C)
C
x=-5x=−5x, equals, minus, 5 and x=7x=7x, equals, 7
(Choice D)
D
x=-5x=−5x, equals, minus, 5 and x=-7x=−7x, equals, minus, 7
[I need help!]
(x+5)(x+7)=0left parenthesis, x, plus, 5, right parenthesis, left parenthesis, x, plus, 7, right parenthesis, equals, 0
\begin{aligned}&\swarrow&\searrow\\\\ x+5&=0&x+7&=0\\\\ x&=-5&x&=-7\end{aligned}
x=-5x, equals, minus, 5x=-7x, equals, minus, 7
Solve (2x-1)(4x-3)=0(2x−1)(4x−3)=0left parenthesis, 2, x, minus, 1, right parenthesis, left parenthesis, 4, x, minus, 3, right parenthesis, equals, 0.
Choose 1 answer:
Choose 1 answer:
(Choice A)
A
x=1x=1x, equals, 1 and x=3x=3x, equals, 3
(Choice B)
B
x=\dfrac{1}{2}x=
2
1
x, equals, start fraction, 1, divided by, 2, end fraction and x=\dfrac{3}{4}x=
4
3
x, equals, start fraction, 3, divided by, 4, end fraction
(Choice C)
C
x=2x=2x, equals, 2 and x=\dfrac{4}{3}x=
3
4
x, equals, start fraction, 4, divided by, 3, end fraction
(Choice D)
D
x=-1x=−1x, equals, minus, 1 and x=-3x=−3x, equals, minus, 3
[I need help!]
(2x-1)(4x-3)=0left parenthesis, 2, x, minus, 1, right parenthesis, left parenthesis, 4, x, minus, 3, right parenthesis, equals, 0
\begin{aligned}&\swarrow&\searrow\\\\ 2x-1&=0&4x-3&=0\\\\ 2x&=1&4x&=3\\\\ x&=\dfrac{1}{2}&x&=\dfrac{3}{4}\end{aligned}
x=\dfrac{1}{2}
x, equals, start fraction, 1, divided by, 2, end fractionx=\dfrac{3}{4}
x, equals, start fraction, 3, divided by, 4, end fraction