factorise : x^2-7x+21
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This eq.has imaginary roots
x²-7x+21=0
D(discriminant)=b²-4ac=(-7)²-4(1)(21)=-35
x=(-b+-√D)/2a
x=(7+-√(-35))/(2×1)
x=(7+-√(35)i)/2
so the factors are (x-(7+√(35)i)/2) and (x-(7-√(35)i)/2)
(x-(7+√(35)i)/2)(x-(7-√(35)i)/2)=x²-7x+21
x²-7x+21=0
D(discriminant)=b²-4ac=(-7)²-4(1)(21)=-35
x=(-b+-√D)/2a
x=(7+-√(-35))/(2×1)
x=(7+-√(35)i)/2
so the factors are (x-(7+√(35)i)/2) and (x-(7-√(35)i)/2)
(x-(7+√(35)i)/2)(x-(7-√(35)i)/2)=x²-7x+21
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