find zeroes of the following polynomials (with steps )
(i) p(x)=6x^2-3-7x
(ii) f(x)=2-x^2
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Answered by
4
(i) 6x² - 7x - 3 = 0
6x² - 9x +2x -3 = 0
3x(2x-3) +1(2x-3) = 0
(2x-3)(3x+1) = 0
x = 3/2 or -1/3
(ii) 2 - x² = 0
x² = 2
x = √2 or -√2
6x² - 9x +2x -3 = 0
3x(2x-3) +1(2x-3) = 0
(2x-3)(3x+1) = 0
x = 3/2 or -1/3
(ii) 2 - x² = 0
x² = 2
x = √2 or -√2
Answered by
2
6x²-7x-3=0
=>6x²-9x-2x-3=0
=>3x(2x-3)-1(2x-3)=0
=>(3x-1)-(2x-3)=0
=>x=1/3 or x=3/2
-x²+2=0
=>+x²-2=0
=>x²=2
=>x=+√2 or x= - √2
=>6x²-9x-2x-3=0
=>3x(2x-3)-1(2x-3)=0
=>(3x-1)-(2x-3)=0
=>x=1/3 or x=3/2
-x²+2=0
=>+x²-2=0
=>x²=2
=>x=+√2 or x= - √2
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