Factorise: - $x^2+7x+6$
- $9x^2-16$
- $(x+6)(x+1)$
- $(3x+4)(3x-4)$
Factorise: - $4x^2-23x+15$
- $-2x^2+9x-4$
- $(4x-3)(x-5)$
- $(1-2x)(x-4) = (2x-1)(4-x)$
A circle has equation $$2x^2+2y^2-16x+12y=0$$ Find the center and the radius of the circle.
Centre $(4,-3)$, radius 5
Solve the following by factorising: - $x^2-3x=10$
- $8x-x^2=15$
Solve the following by completing the square: - $2(x^2-2x)=5$
- $x(x-1)=3$
- $1\pm\sqrt{\dfrac{7}{2}}$
- $\dfrac{1}{2}\pm\dfrac{\sqrt{13}}{2}$
Solve the following using the quadratic equation: - $35=2x+x^2$
- $2x^2+3=6x$
- $x=-7,5$
- $x=\dfrac{3\pm\sqrt{3}}{2}$
Two circles have equations: $$(x-3)^2+(y-2)^2 = 16$$ $$(x+5)^2+y^2=16$$ Determine if they intersect.
Distance between centres: $\sqrt{8^2+2^2}$ <br> Sum of radii is smaller so no intersection.
Solve $$3x^2+(8-3x)^2 = 28$$
$x = 1, 3$
Sketch the following graphs, indicating clearly their points of intersection with the axes and the coordinates of their turning points: - $y=x^2-9$
- $y=2x^2-5x+3$
- $(-3,0)$, $(3,0)$, $(0, -9)$ minimum at $(0,9)$
- $(1,0)$, $(1.5,0)$, $(0, 3)$ minimum at $(\frac{5}{4},-\frac{1}{8})$
A circle has centre $(-1,2)$ and passes through the points $A(-4,3)$ and $B(0,5)$. - Find an equation of the circle.
- Find the equation of the perpendicular bisector of $AB$.
- $(x+1)^2 + (y-2)^2 = 10$
- $y = -2x$
A circle has a diameter between $(8,-7)$ and $(4,5)$. Find an equation of the circle.
$(x-6)^2 + (y+1)^2 = 40$
The function $\mathrm{f}(x) = x^2+3px+14p-3$, where $p$ is an integer, has two equal roots. Find $p$ and solve the equation for this value of $p$.
$p = 6$ and $x = -9$
By completing the square, find the turning points of: - $y=2x^2-x-5$
- $y=2x+9-x^2$
- $\left(\dfrac{1}{4},-\dfrac{41}{8}\right)$
- $(1,10)$
Write the following in the form $a(x+q)^2+q$: - $2x^2+5x+1$
- $7x-5x^2+11$
- $2\left(x+\dfrac{5}{4}\right)^2-\dfrac{17}{8}$
- $-5\left(x-\dfrac{7}{10}\right)^2+\dfrac{269}{20}$
A circle has equation $$x^2 + y^2 - 20x + 8y + 16 = 0$$ - Find the gradient of the line segment joining the centre of the circle to the point $(4,4)$.
- Hence find the equation of the tangent to the circle at the point $(4,4)$.
- $-\dfrac{4}{3}$
- $y=\dfrac{3}{4}x+1$
A circle has centre $(3,-5)$ and equation $x^2 + y^2 - 6x + ay = 15$, where $a$ is a constant. Find the radius of the circle.
$7$
A circle has diameter which has endpoints $(2,5)$ and $(-2,9)$. - Find an equation of the circle.
- Does the point $(1,5)$ lie within the circle?
The points $P(-3,0)$, $Q(-1,6)$ and $R(11,2)$ lie on a circle. - Show that angle $PQR$ is a right angle.
- Find an equation for the circle.
- $PQ$ gradient $3$, $QR$ gradient $-\frac{1}{3}$
- $(x-4)^2+(y-1)^2=50$
By finding the discriminant of $x^2-3x+5$, explain why $(x+2)(x^2-3x+5)$ is always positive for $x > -2$.
The quadratic has no real roots and is always positive. The linear part is also positive when $x > -2$.
Two circles have equations: $(x-7)^2+(y-3)^2=144$ and $(x+2)^2+(y-1)^2=9$. Determine if they intersect.
Distance between centres $\sqrt{9^2+2^2}$ is less than larger radius. <br> Difference between radii is smaller than distance between centres, so intersect at two points.
Find the minimum value of $$(x^2+4x+9)^2$$
$((x+2)^2+5)^2$ has minimum $25$
Given $\mathrm{f}(x) = 2x^2+(k+4)x+k$, where $k$ is a real constant, find the discriminant in terms of $k$ and hence prove that there are two distinct real roots of $\mathrm{f}(x)$ for all values of $k$.
$\Delta = (k+4)^2-8k = k^2+16$. Positive for all values of $k$, so there are always two real roots.
Solve the following: - $x^4-13x^2+36=0$
- $x^6+7x^3=8$
Solve $$x+3=4\sqrt{x}$$
$1, 9$
Solve $$\dfrac{4x^4-24}{10} = x^2$$
$\pm 2$
Solve $$(x^2-4x+1)^2+(x^2-4x+1)=12$$
$2 \pm \sqrt{6}$
Solve $$2^{2x}+2^4=2^{x+1}+2^{x+3}$$
$1, 3$
Solve $$81 +3^{2x+1}=4\times 3^{x+2}$$
$1, 2$
A circle has centre at the origin and radius $r$. It fits completely inside another circle with equation $$x^2 + y^2 - 10x - 24y = 231$$ Find the range of possible values of $r$.
$r < 7$
The function $\mathrm{f}$ is such that $\mathrm{f}(x^n) = \mathrm{f}(x)^n$. Find the possible values of $\mathrm{f}(2)$ given $$2\mathrm{f}(64) - 7\mathrm{f}(16) = 4\mathrm{f}(4)$$
$-\frac{1}{2}, 0, 4$