Show that a^2x^2 - 2ax - a+1 = (ax + squareroot a - 1)(ax - squareroot a -1)
thanks in advance.
Help with Maths (quadratics).?
2009-03-14 10:51 am
回答 (6)
2009-03-14 11:04 am
✔ 最佳答案
a²x² – 2ax – a + 1 = a²x² – 2ax + 1 – a.......................... = (a²x² – 2ax + 1) – (√a)²
.......................... = (ax – 1)² – (√a)²
.......................... = [(ax – 1) + (√a)][(ax – 1) – (√a)]
.......................... = (ax – 1 + √a)(ax – 1 – √a)
.......................... = (ax + √a – 1)(ax – √a – 1)
2009-03-14 11:17 am
Start with the right hand side.
ax +âa-1)(ax-âa-1)
Use the distributive law
=ax(ax-âa-1)+âa(ax-âa-1)-1(ax-âa-1)
Multiply out the brackets
=[a^2x^2 - axâa - ax]+[axâa -a-âa]-[ax-âa-1]
Collect like terms
=a^2x^2 - axâa+axâa-ax-ax -a +1
= a^2x^2 - 2ax - a + 1
ax +âa-1)(ax-âa-1)
Use the distributive law
=ax(ax-âa-1)+âa(ax-âa-1)-1(ax-âa-1)
Multiply out the brackets
=[a^2x^2 - axâa - ax]+[axâa -a-âa]-[ax-âa-1]
Collect like terms
=a^2x^2 - axâa+axâa-ax-ax -a +1
= a^2x^2 - 2ax - a + 1
2009-03-14 11:09 am
a^2x^2 - 2ax - a + 1
= a^2x^2 - 2ax + 1 - a
= a^2x^2 - ax - ax + 1 - a
= (a^2x^2 - ax) - (ax - 1) - a
= ax(ax - 1) - 1(ax - 1) - a
= (ax - 1)(ax - 1) - a
= (ax - 1)^2 - (âa)^2
= [(ax - 1) + âa][(ax - 1) - âa]
= [ax - 1 + âa][ax - 1 - âa]
= [ax + âa - 1][ax - âa - 1]
= a^2x^2 - 2ax + 1 - a
= a^2x^2 - ax - ax + 1 - a
= (a^2x^2 - ax) - (ax - 1) - a
= ax(ax - 1) - 1(ax - 1) - a
= (ax - 1)(ax - 1) - a
= (ax - 1)^2 - (âa)^2
= [(ax - 1) + âa][(ax - 1) - âa]
= [ax - 1 + âa][ax - 1 - âa]
= [ax + âa - 1][ax - âa - 1]
2009-03-14 11:09 am
(m^2-n^2)=(m+n)(m-n)
2009-03-14 11:06 am
a^2x^2 - 2ax - a+1 = a^2x^2 - 2ax - (a - 1)
= a^2x^2 - 2ax - (sqrt(a)^2 - 1)
= ( ax + (sqrt(x) - 1)) (ax - (sqrt(x) + 1))
= (ax + squareroot a - 1)(ax - squareroot a -1)
= a^2x^2 - 2ax - (sqrt(a)^2 - 1)
= ( ax + (sqrt(x) - 1)) (ax - (sqrt(x) + 1))
= (ax + squareroot a - 1)(ax - squareroot a -1)
2009-03-14 10:59 am
You can show this very easily using the cross method to factorize quadratic equations.
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