Quadratics: past paper questions & worked solutions

For no real roots, the discriminant must be negative: $b^2-4ac<0$.

For $kx^2+6kx+5=0$, $a=k$, $b=6k$, and $c=5$.

$(6k)^2-4(k)(5)<0$, so $36k^2-20k<0$.

$4k(9k-5)<0$, so the product is negative when $0<k<\dfrac59$.

Let $u=\sqrt{x}$. Then $x=u^2$, so the equation becomes $u^2-6u+4=0$.

$u=\dfrac{6\pm\sqrt{36-16}}{2}=\dfrac{6\pm2\sqrt5}{2}=3\pm\sqrt5$.

Since $u=\sqrt{x}$, $x=u^2=(3\pm\sqrt5)^2=9+5\pm6\sqrt5=14\pm6\sqrt5$.

$x^2+kx-9=-3x^2-5x+k$ gives $4x^2+(k+5)x-(k+9)=0$.

For a single point of intersection, the discriminant is zero.

$(k+5)^2-4(4)(-(k+9))=0$, so $k^2+26k+169=0$.

$k^2+26k+169=(k+13)^2$, so $k=-13$.

Substituting $k=-13$ gives $4x^2-8x+4=0$, so $4(x-1)^2=0$ and $x=1$.

$y=x^2+kx-9=1-13-9=-21$, so $P=(1,-21)$.

$4(p-2x)=\dfrac{12+15p}{x+p}$ gives $4(p-2x)(x+p)=12+15p$.

Expanding gives $4p^2-4px-8x^2=12+15p$, so $8x^2+4px+12+15p-4p^2=0$.

For two distinct real roots, $b^2-4ac>0$.

$(4p)^2-4(8)(12+15p-4p^2)>0$, so $144p^2-480p-384>0$.

Dividing by 48 gives $3p^2-10p-8>0$, and $3p^2-10p-8=(3p+2)(p-4)$.

The upward-opening quadratic is positive outside its roots, so $p<-\dfrac23$ or $p>4$.

$2x+2y=350$, so $x+y=175$.

$xy=7350$.

Using $y=175-x$, $x(175-x)=7350$, so $x^2-175x+7350=0$.

$x=\dfrac{175\pm\sqrt{175^2-4(7350)}}{2}=\dfrac{175\pm35}{2}$, giving $x=105$ or $70$.

Since $x>y$, $x=105$ and $y=70$.

$3x^2+12x+13=3(x^2+4x)+13$.

$x^2+4x=(x+2)^2-4$, so $3(x^2+4x)+13=3(x+2)^2-12+13$.

$y=3(x+2)^2+1$ has turning point $(-2,1)$. The y-intercept is found from $x=0$, giving $y=13$.

The curve is an upward-opening parabola with minimum $(-2,1)$ and y-intercept $(0,13)$.

$f(x)=11-4x-2x^2=-2(x^2+2x)+11$.

$-2(x^2+2x)+11=-2\big((x+1)^2-1\big)+11=13-2(x+1)^2$.

The graph is a downward-opening parabola with maximum point $(-1,13)$. The y-intercept is $f(0)=11$, so it crosses the y-axis at $(0,11)$.

From $13-2(x+1)^2$, the line of symmetry is $x=-1$.

When $t=3$ and $H=2$, $2^3=9p+q$, so $8=9p+q$. When $t=5$ and $H=2.4$, $2.4^3=25p+q$, so $13.824=25p+q$.

Subtracting the two equations gives $16p=5.824$, so $p=0.364$. Then $8=9(0.364)+q$, giving $q=4.724$, so $q=4.72$ to 3 significant figures.

For $H=5$, $125=0.364T^2+4.724$.

$T^2=\dfrac{125-4.724}{0.364}=330.428\ldots$, so $T=18.177\ldots$. Therefore $T=18.2$ years to one decimal place.

$-2x^2+12x+7=-2(x^2-6x)+7$.

$x^2-6x=(x-3)^2-9$, so $-2x^2+12x+7=-2\big((x-3)^2-9\big)+7=-2(x-3)^2+25$.

Because the coefficient of $(x-3)^2$ is negative, the curve has its maximum when $(x-3)^2=0$. Hence $x=3$, and the maximum value is $y=25$, so $P=(3,25)$.

The line through $O=(0,0)$ and $P=(3,25)$ has gradient $\dfrac{25}{3}$, so $l$ has equation $y=\dfrac{25}{3}x$.

Points in $R$ are below the curve, above the line, and on or to the right of the $y$-axis. Therefore $y\leq-2x^2+12x+7$, $y\geq\dfrac{25}{3}x$, and $x\geq0$.

The equation becomes $3.5x^6+4x^3+0.5=0$.

Multiplying by $2$ gives $7x^6+8x^3+1=0$.

$7x^6+8x^3+1=(x^3+1)(7x^3+1)=0$.

$x^3=-1$ or $x^3=-\dfrac17$, so $x=-1$ or $x=-\sqrt[3]{\dfrac17}$.

Let $u=x^3$. Then the equation is $(k-1)u^2+4u+(k-4)=0$.

Since every real value of $u$ gives a real $x$, the original equation has no real roots exactly when this quadratic in $u$ has no real roots.

Require $b^2-4ac<0$: $16-4(k-1)(k-4)<0$.

$16-4(k^2-5k+4)<0$ gives $20k-4k^2<0$, so $4k(5-k)<0$.

$4k(5-k)<0$ for $k<0$ or $k>5$.

From $y-x^3+6x=0$, we get $y=x^3-6x$.

$2x(x^3-6x)-3x^2=50$, so $2x^4-15x^2-50=0$.

Let $u=x^2$. Then $2u^2-15u-50=0$, giving $u=10$ or $u=-\dfrac52$.

Since $x^2$ cannot be negative, $x^2=10$, so $x=\pm\sqrt{10}$.

$y=x^3-6x=x(x^2-6)=x(10-6)=4x$. Therefore $y=4\sqrt{10}$ when $x=\sqrt{10}$ and $y=-4\sqrt{10}$ when $x=-\sqrt{10}$.

$(x+\sqrt2)^2=x^2+2\sqrt2x+2$.

Since $\sqrt8=2\sqrt2$, $(3x-5\sqrt8)^2=(3x-10\sqrt2)^2=9x^2-60\sqrt2x+200$.

$f(x)=x^2+2\sqrt2x+2+9x^2-60\sqrt2x+200=10x^2-58\sqrt2x+202$.

$\sqrt3(4y-3\sqrt3)=5y+\sqrt3$ gives $4\sqrt3y-9=5y+\sqrt3$.

$y(4\sqrt3-5)=9+\sqrt3$, so $y=\dfrac{9+\sqrt3}{4\sqrt3-5}$.

$y=\dfrac{(9+\sqrt3)(4\sqrt3+5)}{(4\sqrt3)^2-5^2}=\dfrac{57+41\sqrt3}{23}=\dfrac{57}{23}+\dfrac{41}{23}\sqrt3$.

$20x^3-50x^2-30x=10x(2x^2-5x-3)=10x(2x+1)(x-3)$.

$10x(2x+1)(x-3)=0$, so $x=0$, $x=-\frac12$, or $x=3$.

Let $u=(y+3)^{1/2}$. Then the equation becomes $20u^3-50u^2-30u=0$, so $u=0$, $u=-\frac12$, or $u=3$.

Since $u=\sqrt{y+3}\ge0$, reject $u=-\frac12$. If $u=0$, then $y=-3$. If $u=3$, then $y+3=9$, so $y=6$.

The line cuts the y-axis at $-12$ and the x-axis at $4$, so its gradient is $\dfrac{0-(-12)}{4-0}=3$. Hence $l$ is $y=3x-12$.

The minimum point has x-coordinate 7. The roots are $4$ and $k$, so their midpoint is 7.

$\dfrac{4+k}{2}=7$, so $k=10$.

Let $f(x)=a(x-4)(x-10)$. Since the minimum point is $(7,-18)$, $-18=a(3)(-3)=-9a$, so $a=2$.

$f(x)=2(x-4)(x-10)=2x^2-28x+80$.

The region lies between the y-axis and $x=4$, above the line and below the curve: $0\le x\le4$ and $3x-12\le y\le2x^2-28x+80$.

$f(x)=x^3-2x^{5/2}+14$.

$f'(x)=3x^2-5x^{3/2}=x^{3/2}(3x^{1/2}-5)$. Since $x>0$, the factor $x^{3/2}$ is non-zero. Therefore $3\sqrt{x}-5=0$, so $\sqrt{x}=\dfrac53$ and $x=\dfrac{25}{9}$.

$f''(x)=6x-\dfrac{15}{2}x^{1/2}$.

$6x-\dfrac{15}{2}x^{1/2}=66$. Multiplying by 2 and dividing by 3 gives $4x-5x^{1/2}-44=0$, as required.

Let $u=\sqrt{x}$, where $u\geq0$. Then $4u^2-5u-44=0$, and $(u-4)(4u+11)=0$. Hence $u=4$ or $u=-\dfrac{11}{4}$.

Because $u=\sqrt{x}\geq0$, the value $-\dfrac{11}{4}$ is impossible. Thus $\sqrt{x}=4$ and $x=16$, which is the only solution.

$2x^2-16x+50=2(x^2-8x)+50=2[(x-4)^2-16]+50$.

$2x^2-16x+50=2(x-4)^2+18$.

The minimum point is $M=(4,18)$.

Since $l$ passes through $(0,0)$ and $(4,18)$, its gradient is $\dfrac{18}{4}=\dfrac92$, so $l:y=\dfrac92x$.

Set $2x^2-16x+50=\dfrac92x$.

Multiplying by $2$ gives $4x^2-32x+100=9x$, so $4x^2-41x+100=0$.

$4x^2-41x+100=(x-4)(4x-25)=0$. The point $M$ has $x=4$, so $P$ has $x=\dfrac{25}{4}$.

The shaded region is above the line, below the curve, and between the y-axis and $M$.

$\dfrac92x\le y\le2x^2-16x+50$ with $0\le x\le4$.

The maximum is at $(2,20)$ and the curve passes through $(0,0)$. Since the axis of symmetry is $x=2$, the other x-intercept is $(4,0)$.

Let $f(x)=a(x-2)^2+20$.

$0=a(0-2)^2+20$, so $4a=-20$ and $a=-5$.

$f(x)=-5(x-2)^2+20=-5x^2+20x=-5x(x-4)$.

$-5x^2+20x=x(x^2-4)=x^3-4x$.

$x^3+5x^2-24x=0$, so $x(x^2+5x-24)=0$ and $x(x+8)(x-3)=0$.

The negative x-coordinate is $x=-8$.

$y=x(x^2-4)=(-8)(64-4)=-480$, so $P=(-8,-480)$.

$2x^2-12x+14=2(x^2-6x)+14=2\left((x-3)^2-9\right)+14=2(x-3)^2-4$.

The minimum of $2(x-3)^2-4$ is at $P=(3,-4)$.

The line passes through $(-1,28)$ and $P(3,-4)$, so $m=\dfrac{-4-28}{3-(-1)}=\dfrac{-32}{4}=-8$.

Using $P(3,-4)$, $y+4=-8(x-3)$, so $y=-8x+20$.

The region is to the right of the y-axis and above the x-axis, so $x\ge0$ and $y\ge0$.

The region is below the line and above the curve, so $y\le-8x+20$ and $y\ge2x^2-12x+14$.