Quadratic Equations & Polynomials: CAT previous year questions

Solution

Minimum of $f$: at $x = 2c$, $f = 8c - 4c^2$. Maximum of $g$: at $x = \dfrac{3c}{2}$, $g = \dfrac{9c^2}{4} - 2c$.

$8c - 4c^2 > \dfrac{9c^2}{4} - 2c \Rightarrow 10c > \dfrac{25c^2}{4} \Rightarrow 0 < c < 1.6$.

Of the options, only $c = \dfrac{1}{2}$ works.

Solution

Integer roots with sum 5 and product $k \geq 0$: $(0, 5), (1, 4), (2, 3)$.

$k = 0, 4, 6$: $3$ values.

Solution

Let the common root be $\alpha$: $3\alpha^2 - 5\alpha + p = 0$ and $2\alpha^2 - 2\alpha + q = 0$.

$2 \times$ first $- 3 \times$ second: $-4\alpha + 2p - 3q = 0 \Rightarrow \alpha = \dfrac{2p - 3q}{4}$.

Other roots: $\dfrac{5}{3} - \alpha$ and $1 - \alpha$. Sum $= \dfrac{8}{3} - 2\alpha = \dfrac{8}{3} - p + \dfrac{3}{2}q$.

Solution

Let $t = x^2 + 3x$. Then $f(x) + 1 = t(t + 2) + 1 = (t + 1)^2$, so $|t + 1| = 9701$.

$t = 9700$: $x^2 + 3x - 9700 = 0$ has real roots with sum $-3$.

$t = -9702$: $x^2 + 3x + 9702 = 0$ has no real roots.

Sum of real roots $= -3$.

Solution

Subtract the second equation from the first (after dividing both by 4):

$x^2 + y^2 + z^2 - x + y + z = -\dfrac{3}{4}$.

Complete squares: $\left(x - \dfrac{1}{2}\right)^2 + \left(y + \dfrac{1}{2}\right)^2 + \left(z + \dfrac{1}{2}\right)^2 = 0$.

So $x = \dfrac{1}{2}$, $y = z = -\dfrac{1}{2}$, and $a = 4\left(\dfrac{3}{4}\right) = 3$.

Solution

Let the common root be $r$. Adding the first two equations and subtracting the third:

$r^2 + 9 + 17 - 35 = 0 \Rightarrow r^2 = 9 \Rightarrow r = -3$ (negative).

$9 - 3m + 9 = 0 \Rightarrow m = 6$; $9 - 3n + 17 = 0 \Rightarrow n = \dfrac{26}{3}$.

$2m + 3n = 12 + 26 = 38$.

Solution

$\alpha\beta = -\dfrac{1}{3}$. $\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = \dfrac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = 9(\alpha^2 + \beta^2) = 15 \Rightarrow \alpha^2 + \beta^2 = \dfrac{5}{3}$.

$(\alpha + \beta)^2 = \dfrac{5}{3} - \dfrac{2}{3} = 1$.

$\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) = 2(\alpha + \beta)$.

$(\alpha^3 + \beta^3)^2 = 4(\alpha + \beta)^2 = 4$.

Solution

Rewrite: $(2x - y)^2 + 3y^2 - 6y + 3 = (2x - y)^2 + 3(y - 1)^2 = 0$.

Both squares are zero: $y = 1$, $x = \dfrac{1}{2}$.

$4x + 5y = 2 + 5 = 7$.

Solution

Move everything to one side: $x^2 + 4xy + 4y^2 + (x - 2y - 1)^2 = 0$.

$(x + 2y)^2 + (x - 2y - 1)^2 = 0$, so $x - 2y - 1 = 0$.

$x - 2y = 1$.

Solution

Since $-2$ is a root, factor: $(x + 2)(x^2 + (2r - 1)x + 1)$.

The other roots are real if $(2r - 1)^2 \geq 4 \Rightarrow |2r - 1| \geq 2$, i.e. $r \geq 1.5$ or $r \leq -0.5$.

Smallest non-negative integer $r = 2$.

Solution

$\alpha + \beta = 3$, $\alpha\beta = \dfrac{k}{2}$. These are the roots of $x^2 + px + p = 0$:

Sum: $3 + \dfrac{k}{2} = -p$. Product: $\dfrac{3k}{2} = p$.

$3 + \dfrac{k}{2} = -\dfrac{3k}{2} \Rightarrow k = -\dfrac{3}{2}$, $p = -\dfrac{9}{4}$.

$8(k - p) = 8\left(-\dfrac{3}{2} + \dfrac{9}{4}\right) = 6$.

Solution

$(x - 1)^2 + 2kx + 11 = x^2 + (2k - 2)x + 12$. No real roots: $(2k - 2)^2 < 48 \Rightarrow (k - 1)^2 < 12$.

$k - 1 \leq 3$, so the largest integer $k = 4$.

$\dfrac{4}{4y} + 9y = \dfrac{1}{y} + 9y \geq 2\sqrt{9} = 6$ (AM–GM), at $y = \dfrac{1}{3}$.

Least value $= 6$.

Solution

$2pq - 20 = 52 - 2pq \Rightarrow pq = 18$. Then $p^2 + q^2 - 29 = 16 \Rightarrow p^2 + q^2 = 45$.

$(p - q)^2 = 45 - 36 = 9 \Rightarrow p - q = \pm 3$.

$p^3 - q^3 = (p - q)(p^2 + pq + q^2) = \pm 3 \times 63 = \pm 189$.

Difference $= 189 - (-189) = 378$.

Solution

Let $u, v$ be the reciprocals of the roots: $(u - v)^2 = \dfrac{1}{9}$ and $u^2 + v^2 = \dfrac{5}{9}$.

$2uv = \dfrac{4}{9} \Rightarrow uv = \dfrac{2}{9}$; $(u + v)^2 = 1 \Rightarrow u + v = \pm 1$.

So $u, v = \pm\dfrac{1}{3}, \pm\dfrac{2}{3}$, and the roots are $3, \dfrac{3}{2}$ or $-3, -\dfrac{3}{2}$.

Roots $3, 1.5$: $b = -4.5$, $c = 4.5$, $b + c = 0$. Roots $-3, -1.5$: $b = 4.5$, $c = 4.5$, $b + c = 9$.

Largest $= 9$.

Solution

$b^2 < 4ac$ means $f(x)$ has no real roots, so $f(x)$ has the same sign as $a$ for every $x$.

If $a > 0$, $f(m) > 0$ for all $m$: S is empty. If $a < 0$, $f(m) < 0$ for all $m$: S is all integers.

So S is either the empty set or the set of all integers.

Solution

$a(a + b + 1) = 14$ and $b(a + b + 1) = 28$.

Adding: $(a + b)(a + b + 1) = 42 = 6 \times 7 \Rightarrow a + b = 6$.

Then $7a = 14 \Rightarrow a = 2$, $b = 4$. $2a + b = 8$.

Solution

$A^B = 1$ when $A = 1$, or $B = 0$ (with $A \neq 0$), or $A = -1$ with $B$ even.

$x^2 - 10 = 1$: no integer $x$.

$x^2 - 3x - 10 = 0$: $x = 5$ or $-2$ (bases 15 and $-6$) ✓.

$x^2 - 10 = -1$: $x = \pm 3$; exponents $-10$ and $8$ are even ✓.

Total $= 4$ solutions.

Solution

$f(x) \geq 0$ everywhere and $f(2) = 0$, so 2 is a double root: $f(x) = a(x - 2)^2$.

$f(4) = 4a = 6 \Rightarrow a = 1.5$.

$f(-2) = 1.5 \times 16 = 24$.

Solution

Let the roots be $r, -r, s$.

Sum of pairwise products: $-r^2 + rs - rs = -r^2 = -\dfrac{10}{5} \Rightarrow r^2 = 2$.

Product: $-r^2 s = -\dfrac{9}{5} \Rightarrow s = \dfrac{9}{10}$.

Sum of roots: $s = -\dfrac{c}{5} \Rightarrow c = -\dfrac{9}{2}$.

Solution

$2x^2 + kx + 5$ has no real roots: $k^2 < 40 \Rightarrow -6 \leq k \leq 6$.

$x^2 + (k - 5)x + 1$ has two distinct real roots: $(k - 5)^2 > 4 \Rightarrow k < 3$ or $k > 7$.

Together: $k = -6, -5, \dots, 2$: $9$ values.

Solution

With integer coefficients, $3 - 2\sqrt{2}$ is also a root: sum $= 6$, product $= 1$, so $b = -6a$, $c = a$.

Similarly $4 \pm 2\sqrt{3}$: sum $= 8$, product $= 4$, so $m = -8a$, $n = 4a$.

$\dfrac{b}{m} + \dfrac{c - 2b}{n} = \dfrac{6}{8} + \dfrac{13a}{4a} = \dfrac{3}{4} + \dfrac{13}{4} = 4$.

Solution

$x^2 - 4x - 13 = (x - 2)^2 - 17$, with minimum $-17$.

$|g(x)| = r$ means $g(x) = r$ or $g(x) = -r$. $g(x) = r$ always gives two roots (for $r > 0$).

For exactly three roots, $g(x) = -r$ must give one root: $-r = -17 \Rightarrow r = 17$.

Solution

Coefficients are rational, so $2 - \sqrt{3}$ is the other root: sum $= 4$, product $= 1$.

$\dfrac{b}{a} = 4$ and $\dfrac{c}{a} = 1$, so $b = 4a$, $c = a$.

$b = c^3 \Rightarrow 4a = a^3 \Rightarrow a^2 = 4 \Rightarrow |a| = 2$.

Solution

Write $d = x - y$. First: $x(d - 1) = 22$. Second: $y(1 - d) = 34$, i.e. $-y(d - 1) = 34$.

Adding: $(x - y)(d - 1) = 56 \Rightarrow d(d - 1) = 56$.

$d = 8$ (as $x > y$).

Solution

Left side: $2^x + 2^{-x} \geq 2$, equal to 2 only at $x = 0$.

Right side: $2 - (x - 2)^2 \leq 2$, equal to 2 only at $x = 2$.

They can only be equal if both are 2, which needs $x = 0$ and $x = 2$ at once. No real solutions: $0$.

Solution

$A^B = 1$: base 1, exponent 0, or base $-1$ with even exponent.

Base 1: $x^2 - 7x + 10 = 0 \Rightarrow x = 2, 5$.

Exponent 0: $x^2 - 13x + 42 = 0 \Rightarrow x = 6, 7$ (bases 5 and 11, non-zero) ✓.

Base $-1$: $x^2 - 7x + 12 = 0 \Rightarrow x = 3, 4$; exponents 12 and 6 are even ✓.

Total $= 6$.

Solution

Let $t = x + \dfrac{1}{x}$: $t^2 - 3t + 2 = 0 \Rightarrow t = 1$ or $2$.

$x + \dfrac{1}{x} = 1$ has no real solution (for real $x$, $|t| \geq 2$).

$x + \dfrac{1}{x} = 2 \Rightarrow x = 1$.

Number of distinct real roots $= 1$.

Solution

Base 1: $x^2 - 5x + 6 = 0 \Rightarrow x = 2, 3$.

Exponent 0: $x = -1$ (base $= 13 \neq 0$) ✓.

Base $-1$: $x^2 - 5x + 8 = 0$ has no real roots.

Total $= 3$ integers.

Solution

$|a - 2| = 2|x| - x^2 \geq 0 \Rightarrow |x| \leq 2$.

$x = 0$: $|a - 2| = 0 \Rightarrow a = 2$ (1 pair).

$x = \pm 1$: $|a - 2| = 1 \Rightarrow a = 1, 3$ (4 pairs).

$x = \pm 2$: $|a - 2| = 0 \Rightarrow a = 2$ (2 pairs).

Total $= 7$.

Solution

Real roots: $m^2 \geq 8n$ and $4n^2 \geq 4m \Rightarrow n^2 \geq m$.

$n = 1$: $m \leq 1$ but $m^2 \geq 8$ ✗.

$n = 2$: $m \leq 4$ and $m^2 \geq 16 \Rightarrow m = 4$; $m + n = 6$.

Larger $n$ gives a larger sum (e.g. $n = 3$ needs $m \geq 5$). Smallest $m + n = 6$.