Question

soove by factoristion method x²-x-a (a + 1) = 0​

Answer 1

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Yes! When the x2 coefficient is 1 , we call this a monic quadratic, in this case. The coefficient of x, the minus one, here, is the negative of the sum of the roots. The constant term is the product of the roots.

If the roots are named p and q, then we have: p+q= 1 and pq=-a• (a+1). Obviously, (a+1) -a = 1, so -a and (a+1) are the two roots.

You can also see this by subtracting the constant term from both sides and multiplying out the r.h.s or factoring the l.h.s. and identifying the parts in relation to the 1′s and x's. Like so:

x2 − x = a2 + a , -a = x;

x∙(x−1) = (a+1)∙a , (a+1) = x.

Whenever you use an unconventional technique like this one, you should always check that your answer is complete and valid. So we have -a and a+1 as roots. They are distinct unless, a+1 = -a, or a = -1/2. When a =-1/2 both roots are equal ( double ). All other values of “a" give distinct roots.

Also:

(x− −a)∙(x− (a+1)) = (x+a)∙(x−a−1) = x2−ax−x + ax −a(a+1) = x2 − x − a∙(a+1)

So, the two roots produce the equation. In checking, sometimes you will have a nonzero constant, other than one, multiplying the equation. So long as you can legally cancel the extra multiplier, the check is valid.

Answer 2

Answer:

Yes! When the x2 coefficient is 1 , we call this a monic quadratic, in this case. The coefficient of x, the minus one, here, is the negative of the sum of the roots. The constant term is the product of the roots.

If the roots are named p and q, then we have: p+q= 1 and pq=-a• (a+1). Obviously, (a+1) -a = 1, so -a and (a+1) are the two roots.

You can also see this by subtracting the constant term from both sides and multiplying out the r.h.s or factoring the l.h.s. and identifying the parts in relation to the 1′s and x's. Like so:

x2 − x = a2 + a , -a = x;

x∙(x−1) = (a+1)∙a , (a+1) = x.

Whenever you use an unconventional technique like this one, you should always check that your answer is complete and valid. So we have -a and a+1 as roots. They are distinct unless, a+1 = -a, or a = -1/2. When a =-1/2 both roots are equal ( double ). All other values of “a" give distinct roots.

Also:

(x− −a)∙(x− (a+1)) = (x+a)∙(x−a−1) = x2−ax−x + ax −a(a+1) = x2 − x − a∙(a+1)

So, the two roots produce the equation. In checking, sometimes you will have a nonzero constant, other than one, multiplying the equation. So long as you can legally cancel the extra multiplier, the check is valid.