1. `3x^2+6x-1`, find Zeros of a polynomial
Solution:
The Given Polynomial `=3x^2+6x-1`
`3x^2+6x-1=0`
`=>3x^2+6x-1 = 0`
factor is not possible for equation `3x^2+6x-1=0`
But we are trying find solution using the method of perfect square.
Comparing the given equation with the standard quadratic equation `ax^2+bx+c=0,`
we get, `a=3, b=6, c=-1.`
`:. Delta=b^2-4ac`
`=(6)^2-4 (3) (-1)`
`=36+12`
`=48`
`:. sqrt(Delta)=sqrt(48)=4sqrt(3)`
Now, `alpha=(-b+sqrt(Delta))/(2a)`
`=(-(6)+4sqrt(3))/(2*3)`
`=(-6+4sqrt(3))/6`
`=(-3+2sqrt(3))/3`
and, `beta=(-b-sqrt(Delta))/(2a)`
`=(-(6)-4sqrt(3))/(2*3)`
`=(-6-4sqrt(3))/6`
`=(-3-2sqrt(3))/3`
`=>x = (-3+2sqrt(3))/3" or "x = (-3-2sqrt(3))/3`
2. `x^2+3x-4`, find Zeros of a polynomial
Solution:
The Given Polynomial `=x^2+3x-4`
`x^2+3x-4=0`
`=>x^2+3x-4 = 0`
`=>x^2-x+4x-4 = 0`
`=>x(x-1)+4(x-1) = 0`
`=>(x-1)(x+4) = 0`
`=>(x-1) = 0" or "(x+4) = 0`
`=>x = 1" or "x = -4`
3. `2x^2-3x+1`, find Zeros of a polynomial
Solution:
The Given Polynomial `=2x^2-3x+1`
`2x^2-3x+1=0`
`=>2x^2-3x+1 = 0`
`=>2x^2-x-2x+1 = 0`
`=>x(2x-1)-1(2x-1) = 0`
`=>(2x-1)(x-1) = 0`
`=>(2x-1) = 0" or "(x-1) = 0`
`=>2x = 1" or "x = 1`
`=>x = 1/2" or "x = 1`
4. `x^3-2x^2-x+2`, find Zeros of a polynomial
Solution:
The Given Polynomial `=x^3-2x^2-x+2`
`x^3-2x^2-x+2=0`
`=>x^3-2x^2-x+2 = 0`
`=>x^3-x^2-x^2+x-2x+2 = 0`
`=>x^2(x-1)-x(x-1)-2(x-1) = 0`
`=>(x-1)(x^2-x-2) = 0`
`=>(x-1)(x^2+x-2x-2) = 0`
`=>(x-1)(x(x+1)-2(x+1)) = 0`
`=>(x-1)(x+1)(x-2) = 0`
`=>(x-1) = 0" or "(x+1) = 0" or "(x-2) = 0`
`=>x = 1" or "x = -1" or "x = 2`
This material is intended as a summary. Use your textbook for detail explanation.
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