How to solve Quadratic Equations with an Unknown C and other variables.

Question

For instance $3x^2 - 11x + r$, I understand the value of $r$ is $6$ through trial and error but trial and error is extremely inefficient and time consuming thus not useful in exam situations, How would I solve this?

And also : If $x^2 +px + q$ is a factor of $x^3 + ax^2 + b$,

prove that:

$b = q(a-p)$

Answer 1

For your second question, long division gives us: $$ \require{enclose} \begin{array}{rll} x + k \phantom{0000000000i} && \\[-3pt] x^2+px+q \enclose{longdiv}{x^3+ax^2+0x+b}\kern-.2ex \\[-3pt] \underline{x^3+px^2+qx\phantom{00ii}} && \\[-3pt] \phantom{0} kx^2-qx+b \phantom{0000i}&& \hbox{(set $k=a-p$)} && \\[-3pt] \underline{kx^2+kpx+kq}\phantom{0ii} && \\[-3pt] -(q+kp)x+b-kq && \\[-3pt] \ \\[-3pt] \phantom{00} \end{array} $$ So we expect $-(q+kp)x+b-kq$ to be $0$. Therefore we've established that $$(x^2+px+q)(x+a-p)=x^3+ax^2+b.$$ In this arrangement, $b$ is clearly the result of multiplication of the constants, i.e. $b = qa - qp = q(a-p)$.

Answer 2

A: $3x^2-11x +r=0$

The codition is that discreminet $\Delta\geqslant 0$, so we must have:

$\Delta= 11^2-12 r\geqslant0$

Or: $r\leqslant 10$

Now $121$ is odd and $12r$ is even and we have to find odd perfect squares less than $121$ which are $81, 49, 25$ and $9$. Only $49$ and $25$ give integers for $r$:

$r= 121-49=72=12r$ so $r=6$

$r=121-25=96=12r$ so $r=8$

B: $x^2+px+q|x^3+ax+b$

By direct division you find that thr remainder is $(a-p)x^2-qx+b$ which must be equal to zero, so we have:

$x[(a-p)x-q]=-b$

Therefore b must divide both $q$ and $(a-p)$ because $(a-p)$ and $q$ have no common divisors, hence we must have $b|q(a-p)$

Answer 3

The value of r could be anything, but I'm assuming that you mean r should be an integer and the quadratic should be factorable.

In that case, you can use the factoring to reverse engineer the quadratic. Let's say

$(3x+u)(x+w) = 3x^2 -11x +r$

where u and w are integers. Because of how multiplication of binomials works, the sum of u and w equals -11. In addition, they also are the factors of r. That being said, you can set up a system of equations to solve for r.

$u + 3w = -11$

$uw = r$

From there, some guessing and checking is necessary, but if you know your times tables, it shouldn't be too bad. Questions like these can definitely be tedious if they do not provide a point that exists on the curve.

As for the second part of your question, you can multiply the quadratic expression out with a generic binomial to organize your work. You should be able to see that for coefficient of x to the first power to equal zero in the cubic function, some measures must be taken. Let w be an integer.

$(x+w)(x^2 + px + q) = x^3 + ax^2 + b$

$x^3 + px^2 + wx^2 + qx + wpx + qw = x^3 + ax^2 + b$

For the coefficient of x to be zero in the cubic, the sum of q and wp must equal zero. However, w isn't in the cubic. We can find it, though. Let's simplify the equation first.

$px^2 + wx^2 + (q + wp)x + qw = ax^2 +b$

From here, we can already see that $p + w = a$ and $qw = b$. Therefore, we can substitute as follows.

$w = b/q$

$a = (p + b/q)$

Therefore, after canceling out the constants:

$(p + b/q)x^2 + (q + bp/q)x = ax^2

Now solve for b.

$a = p + b/q$

$b/q = a - p$

$b = q(a - p)$

Sorry for the long response and if it doesn't explain the material the way you wanted. I don't come on the stack as often as I used to, so I'm a bit out of practice.