Emma

August 24, 2020

Chapters

 

Solve the Biquadratic Equations

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First Lesson Free>

Exercise 1

{ x }^{ 4 }-10{ x }^{ 2 }+9=0

Exercise 2

{ x }^{ 4 }-13{ x }^{ 2 }+36=0

Exercise 3

{ x }^{ 4 }-61{ x }^{ 2 }+900=0

Exercise 4

{ x }^{ 4 }-25{ x }^{ 2 }+144=0

Exercise 5

{ x }^{ 4 }-16{ x }^{ 2 }-225=0

Exercise 6

{ x }^{ 6 }-7{ x }^{ 3 }+6=0

 

 

Solution of exercise 1

{ x }^{ 4 }-10{ x }^{ 2 }+9=0

Let { x }^{ 2 }=t:

{ t }^{ 2 }-10t+9=0

t=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

t=\frac { -(-10)\pm \sqrt { { (-10) }^{ 2 }-4(1)(9) } }{ 2(1) }

t=\frac { 10\pm \sqrt { 100-36 } }{ 2 }

t=\frac { 10\pm \sqrt { 64 } }{ 2 }

t=\frac { 10\pm 8 }{ 2 }

t=\frac { 10 + 8 }{ 2 } \qquad t=\frac { 10 - 8 }{ 2 }

t=\frac { 18 }{ 2 } \qquad t=\frac { 2 }{ 2 }

t=9 \qquad t=1

{ x }^{ 2 }=9 \qquad { x }^{ 2 }=1

x=\pm3 \qquad x=\pm1

Solution of exercise 2

{ x }^{ 4 }-13{ x }^{ 2 }+36=0

Let { x }^{ 2 }=t:

{ t }^{ 2 }-13t+36=0

t=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

t=\frac { -(-13)\pm \sqrt { { (-13) }^{ 2 }-4(1)(36) } }{ 2(1) }

t=\frac { 13\pm \sqrt { 169-144 } }{ 2 }

t=\frac { 13\pm \sqrt { 25 } }{ 2 }

t=\frac { 13\pm 5 }{ 2 }

t=\frac { 13 + 5 }{ 2 } \qquad t=\frac { 13 - 5 }{ 2 }

t=\frac { 18 }{ 2 } \qquad t=\frac { 8 }{ 2 }

t=9 \qquad t=4

{ x }^{ 2 }=9 \qquad { x }^{ 2 }=4

x=\pm 3 \qquad x=\pm 2

 

Solution of exercise 3

{ x }^{ 4 }-61{ x }^{ 2 }+900=0

 

Let { x }^{ 2 }=t:

{ t }^{ 2 }-61t+900=0

t=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

t=\frac { -(-61)\pm \sqrt { { (-61) }^{ 2 }-4(1)(900) } }{ 2(1) }

t=\frac { 61\pm \sqrt { 3721-3600 } }{ 2 }

t=\frac { 61\pm \sqrt { 121 } }{ 2 }

t=\frac { 61\pm 11 }{ 2 }

t=\frac { 61 + 11 }{ 2 } \qquad t=\frac { 61 - 11 }{ 2 }

t=\frac { 72 }{ 2 } \qquad t=\frac { 50 }{ 2 }

t=36 \qquad t=25

{ x }^{ 2 }=36 \qquad { x }^{ 2 }=25

x= \pm 6 \qquad x= \pm 5

 

Solution of exercise 4

{ x }^{ 4 }-25{ x }^{ 2 }+144=0

Let { x }^{ 2 }=t:

{ t }^{ 2 }-25t+144=0

t=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

t=\frac { -(-25)\pm \sqrt { { (-25) }^{ 2 }-4(1)(144) } }{ 2(1) }

t=\frac { 25\pm \sqrt { 625-576 } }{ 2 }

t=\frac { 25\pm \sqrt { 49 } }{ 2 }

t=\frac { 25\pm 7 }{ 2 }

t=\frac { 25 + 7 }{ 2 } \qquad t=\frac { 25 - 7 }{ 2 }

t=\frac { 32 }{ 2 } \qquad t=\frac { 18 }{ 2 }

t=16 \qquad t=9

{ x }^{ 2 } = 16 \qquad { x }^{ 2 }=9

x = \pm 4 \qquad x = \pm 3

 

Solution of exercise 5

{ x }^{ 4 }-16{ x }^{ 2 }-225=0

Let { x }^{ 2 }=t:

{ t }^{ 2 }-16t-225=0

t=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

t=\frac { -(-16)\pm \sqrt { { (-16) }^{ 2 }-4(1)(-225) } }{ 2(1) }

t=\frac { 16\pm \sqrt { 256+900 } }{ 2 }

t=\frac { 16\pm \sqrt { 1156 } }{ 2 }

t=\frac { 16\pm 34 }{ 2 }

t=\frac { 16 + 34 }{ 2 } \qquad t=\frac { 16 - 34 }{ 2 }

t=\frac { 50 }{ 2 } \qquad t=\frac { -18 }{ 2 }

t=25 \qquad t=-9

{ x }^{ 2 } = 25 \qquad { x }^{ 2 } = -9

x= \pm 5 \qquad x= \pm \sqrt { -9 }\notin R

Solution of exercise 6

{ x }^{ 6 }-7{ x }^{ 3 }+6=0

Let { x }^{ 3 }=t:

{ t }^{ 2 }-7t+6=0

t=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }

t=\frac { -(-7)\pm \sqrt { { (-7) }^{ 2 }-4(1)(6) } }{ 2(1) }

t=\frac { 7\pm \sqrt { 49-24 } }{ 2 }

t=\frac { 7\pm \sqrt { 25 } }{ 2 }

t=\frac { 7\pm 5 }{ 2 }

t=\frac { 7 + 5 }{ 2 } \qquad t=\frac { 7 - 5 }{ 2 }

t=\frac { 12 }{ 2 } \qquad t=\frac { 2 }{ 2 }

t=6 \qquad t=1

{ x }^{ 3 }=6 \qquad { x }^{ 3 }=1

x=\sqrt [ 3 ]{ 6 } \qquad x=1

 
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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.