Quadratic Equation flash cards

1. 1.For $ax^2+bx+c=0$ with roots $\alpha,\beta$, write the sum and product of roots.

   $\alpha+\beta=-\dfrac{b}{a}$ and $\alpha\beta=\dfrac{c}{a}$.

   Hint: Compare coefficients with $a(x-\alpha)(x-\beta)$.

2. 2.What is the discriminant $D$ of $ax^2+bx+c=0$ and what do its signs indicate for real coefficients?

   $D=b^2-4ac$: if $D>0$ roots are real & distinct, $D=0$ roots are real & equal, $D<0$ roots are complex conjugates.

   Hint: Comes from the quadratic formula $x=\dfrac{-b\pm\sqrt{D}}{2a}$.

3. 3.If $\alpha,\beta$ are roots of $ax^2+bx+c=0$, what equation has roots $\dfrac{1}{\alpha},\dfrac{1}{\beta}$?

   $cx^2+bx+a=0$ (i.e., coefficients reversed), valid when $c\neq0$.

   Hint: Substitute $x\to\dfrac{1}{x}$ in the original equation.

4. 4.State the condition for $ax^2+bx+c=0$ ($a,b,c$ rational) to have irrational conjugate roots $p\pm\sqrt{q}$.

   If one irrational root is $p+\sqrt{q}$, the other must be $p-\sqrt{q}$ (irrational roots occur in conjugate pairs when coefficients are rational).

   Hint: Analogous to the complex conjugate root theorem.

5. 5.For $f(x)=ax^2+bx+c$ with $a>0$, what is the minimum value of $f(x)$ and at which $x$ does it occur?

   Minimum value $=-\dfrac{D}{4a}=c-\dfrac{b^2}{4a}$, occurring at $x=-\dfrac{b}{2a}$.

   Hint: Complete the square: $a\left(x+\dfrac{b}{2a}\right)^2-\dfrac{D}{4a}$.

6. 6.For what values of $m$ do both roots of $x^2-2mx+m^2-1=0$ lie strictly between $-2$ and $4$?

   The roots are $x=\dfrac{2m\pm\sqrt{4}}{2}=m\pm1$. We need $-2-1$ and $m<3$. So $m\in(-1,3)$.

   Hint: The discriminant here is a perfect square — solve for the roots directly.

7. 7.If both roots of $x^2-(k+1)x+k^2+k-8=0$ are less than $2$, find the set of values of $k$.

   Three conditions are needed with $f(x)=x^2-(k+1)x+(k^2+k-8)$: (i) $D\ge0 \Rightarrow -3k^2-2k+33\ge0 \Rightarrow k\in[-11/3,3]$; (ii) $f(2)>0 \Rightarrow k^2-k-6>0 \Rightarrow k<-2$ or $k>3$; (iii) vertex $\dfrac{k+1}{2}<2\Rightarrow k<3$. Intersecting all three gives $k\in\left[-\dfrac{11}{3},-2\right)$.

   Hint: Use the standard three-condition test: discriminant, $f(2)$ sign, vertex position.

8. 8.For what values of $a$ does exactly one root of $x^2+(a-1)x+(a^2-5a+6)=0$ lie in the open interval $(0,2)$?

   Exactly one root lies in $(0,2)$ iff $f(0)\cdot f(2)<0$. Here $f(0)=(a-2)(a-3)$ and $f(2)=a^2-3a+8$, whose own discriminant $9-32<0$ makes it always positive. So the condition reduces to $f(0)<0$, i.e. $(a-2)(a-3)<0$, giving $a\in(2,3)$.

   Hint: Check the sign of $f(2)$ first — it may always be positive.

9. 9.If one root of $x^2-(a+1)x+(a^2+a-8)=0$ exceeds $2$ while the other is less than $2$, find the range of $a$.

   Since $2$ lies strictly between the roots and the leading coefficient is positive, we need $f(2)<0$. Computing, $f(2)=a^2-a-6=(a-3)(a+2)<0$, so $a\in(-2,3)$.

   Hint: A number between two roots of an upward parabola makes $f$ negative there.

10. 10.Find all real $a$ for which both roots of $x^2-2(a-1)x+a(a-3)=0$ are positive.

    We need $D\ge0$, sum $>0$, product $>0$. $D/4=(a-1)^2-a(a-3)=a+1\ge0\Rightarrow a\ge-1$. Sum $=2(a-1)>0\Rightarrow a>1$. Product $=a(a-3)>0\Rightarrow a<0$ or $a>3$. Intersecting all three (with $a>1$) forces $a\in(3,\infty)$.

    Hint: Both roots positive needs sum $>0$ and product $>0$, plus real roots.

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