Question

Find all solutions to the following quadratic equations, and write each equation in factored form.

a. x^2 + 25 = 0
b. −x^2 − 16 = −7
c. (x + 2)^2 + 1 = 0
d. (x + 2)^2 = x
e. (x^2 + 1)^2 + 2(x^2 + 1) − 8 = 0
f. (2x − 1)^2 = (x + 1)^2 − 3
g. x^3 + x^2 − 2x = 0
h. x^3 − 2x^2 + 4x − 8 = 0

Answer 1

(a) The solutions are:
`x=5i,\:x=-5i`

(b) The solutions are:
`x=3i,\:x=-3i`

(c) The solutions are:
`x=i-2,\:x=-i-2`

(d) The solutions are:
`x=-(3)/(2)+i(√(7))/(2),\:x=-(3)/(2)-i(√(7))/(2)`

(e) The solutions are:
`x=1,\:x=-1,\:x=√(5)i,\:x=-√(5)i`

(f) The solutions are:
`x=1`

(g) The solutions are:
`x=0,\:x=1,\:x=-2`

(h) The solutions are:
`x=2,\:x=2i,\:x=-2i`

Explanation:

To find the solutions of these quadratic equations you must:

(a) For
`x^2+25=0`

`\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\x^2+25-25=0-25`

`\mathrm{Simplify}\\x^2=-25`

`\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=√(f\left(a\right)),\:\:-√(f\left(a\right))\\\\x=√(-25),\:x=-√(-25)`

`\mathrm{Simplify}\:√(-25)\\\\\mathrm{Apply\:radical\:rule}:\quad √(-a)=√(-1)√(a)\\\\√(-25)=√(-1)√(25)\\\\\mathrm{Apply\:imaginary\:number\:rule}:\quad √(-1)=i\\\\√(-25)=√(25)i\\\\√(-25)=5i`

`-√(-25)=-5i`

The solutions are:
`x=5i,\:x=-5i`

(b) For
`-x^2-16=-7`

`-x^2-16+16=-7+16\\-x^2=9\\(-x^2)/(-1)=(9)/(-1)\\x^2=-9\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=√(f\left(a\right)),\:\:-√(f\left(a\right))\\x=√(-9),\:x=-√(-9)`

The solutions are:
`x=3i,\:x=-3i`

(c) For
`\left(x+2\right)^2+1=0`

`\left(x+2\right)^2+1-1=0-1\\\left(x+2\right)^2=-1\\\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=√(f\left(a\right)),\:\:-√(f\left(a\right))\\\\x+2=√(-1)\\x+2=i\\x=i-2\\\\x+2=-√(-1)\\x+2=-i\\x=-i-2`

The solutions are:
`x=i-2,\:x=-i-2`

(d) For
`\left(x+2\right)^2=x`

`\mathrm{Expand\:}\left(x+2\right)^2= x^2+4x+4`

`x^2+4x+4=x\\x^2+4x+4-x=x-x\\x^2+3x+4=0`

For a quadratic equation of the form
`ax^2+bx+c=0` the solutions are:

`x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)`

`\mathrm{For\:}\quad a=1,\:b=3,\:c=4:\quad x_(1,\:2)=(-3\pm √(3^2-4\cdot \:1\cdot \:4))/(2\cdot \:1)`

`x_1=(-3+√(3^2-4\cdot \:1\cdot \:4))/(2\cdot \:1)=\quad -(3)/(2)+i(√(7))/(2)\\\\x_2=(-3-√(3^2-4\cdot \:1\cdot \:4))/(2\cdot \:1)=\quad -(3)/(2)-i(√(7))/(2)`

The solutions are:
`x=-(3)/(2)+i(√(7))/(2),\:x=-(3)/(2)-i(√(7))/(2)`

(e) For
`\left(x^2+1\right)^2+2\left(x^2+1\right)-8=0`

`\left(x^2+1\right)^2= x^4+2x^2+1\\\\2\left(x^2+1\right)= 2x^2+2\\\\x^4+2x^2+1+2x^2+2-8\\x^4+4x^2-5`

`\mathrm{Rewrite\:the\:equation\:with\:}u=x^2\mathrm{\:and\:}u^2=x^4\\u^2+4u-5=0\\\\\mathrm{Solve\:with\:the\:quadratic\:equation}\:u^2+4u-5=0`

`u_1=(-4+√(4^2-4\cdot \:1\left(-5\right)))/(2\cdot \:1)=\quad 1\\\\u_2=(-4-√(4^2-4\cdot \:1\left(-5\right)))/(2\cdot \:1)=\quad -5`

`\mathrm{Substitute\:back}\:u=x^2,\:\mathrm{solve\:for}\:x\\\\\mathrm{Solve\:}\:x^2=1=\quad x=1,\:x=-1\\\\\mathrm{Solve\:}\:x^2=-5=\quad x=√(5)i,\:x=-√(5)i`

The solutions are:
`x=1,\:x=-1,\:x=√(5)i,\:x=-√(5)i`

(f) For
`\left(2x-1\right)^2=\left(x+1\right)^2-3`

`\left(2x-1\right)^2=\quad 4x^2-4x+1\\\left(x+1\right)^2-3=\quad x^2+2x-2\\\\4x^2-4x+1=x^2+2x-2\\4x^2-4x+1+2=x^2+2x-2+2\\4x^2-4x+3=x^2+2x\\4x^2-4x+3-2x=x^2+2x-2x\\4x^2-6x+3=x^2\\4x^2-6x+3-x^2=x^2-x^2\\3x^2-6x+3=0`

`\mathrm{For\:}\quad a=3,\:b=-6,\:c=3:\quad x_(1,\:2)=(-\left(-6\right)\pm √(\left(-6\right)^2-4\cdot \:3\cdot \:3))/(2\cdot \:3)\\\\x_(1,\:2)=(-\left(-6\right)\pm √(0))/(2\cdot \:3)\\x=(-\left(-6\right))/(2\cdot \:3)\\x=1`

The solutions are:
`x=1`

(g) For
`x^3+x^2-2x=0`

`x^3+x^2-2x=x\left(x^2+x-2\right)\\\\x^2+x-2:\quad \left(x-1\right)\left(x+2\right)\\\\x^3+x^2-2x=x\left(x-1\right)\left(x+2\right)=0`

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

`x=0\\x-1=0:\quad x=1\\x+2=0:\quad x=-2`

The solutions are:
`x=0,\:x=1,\:x=-2`

(h) For
`x^3-2x^2+4x-8=0`

`x^3-2x^2+4x-8=\left(x^3-2x^2\right)+\left(4x-8\right)\\x^3-2x^2+4x-8=x^2\left(x-2\right)+4\left(x-2\right)\\x^3-2x^2+4x-8=\left(x-2\right)\left(x^2+4\right)`

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

`x-2=0:\quad x=2\\x^2+4=0:\quad x=2i,\:x=-2i`

The solutions are:
`x=2,\:x=2i,\:x=-2i`