Building quadratic equations with prescribed roots

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Build a quadratic equation that has:

• (a) one positive and one negative solution;
• (b) two positive solutions;
• (c) two negative solutions.

In each case check that the solutions exist.

Solution

We use Vieta’s formulas: if a monic equation $x^2-s\,x+p=0$ has roots $x_1,x_2$, then $s=x_1+x_2$ (sum) and $p=x_1 x_2$ (product). The signs of the roots are controlled exactly by $s$ and $p$:

• product $p<0$ $\Rightarrow$ roots of opposite sign (one positive, one negative);
• product $p>0$ and sum $s>0$ $\Rightarrow$ both positive;
• product $p>0$ and sum $s<0$ $\Rightarrow$ both negative.

The existence of two real roots requires in every case $\Delta=s^2-4p\ge0$ (with $\Delta>0$ for distinct roots).

(a) One positive and one negative. Choose the roots $x_1=2$, $x_2=-3$. $s=2+(-3)=-1,\quad p=2\cdot(-3)=-6 \;\Rightarrow\; x^2+x-6=0.$ $\Delta=1^2-4\cdot1\cdot(-6)=25>0.$ The roots exist; since $p=-6<0$ they have opposite signs, as required.

(b) Two positive. Choose the roots $x_1=2$, $x_2=3$. $s=5,\quad p=6 \;\Rightarrow\; x^2-5x+6=0.$ $\Delta=(-5)^2-4\cdot6=25-24=1>0.$ The roots exist; $p=6>0$ and $s=5>0$, so both are positive.

(c) Two negative. Choose the roots $x_1=-2$, $x_2=-3$. $s=-5,\quad p=6 \;\Rightarrow\; x^2+5x+6=0.$ $\Delta=5^2-4\cdot6=25-24=1>0.$ The roots exist; $p=6>0$ and $s=-5<0$, so both are negative. $\boxed{\text{(a) } x^2+x-6=0 \qquad \text{(b) } x^2-5x+6=0 \qquad \text{(c) } x^2+5x+6=0}$