Equations, formulae and identities flashcards

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Expand and simplify (x + 3)(x − 5).
\(x^2-5x+3x-15=x^2-2x-15\). Multiply every term in one bracket by every term in the other, then collect like terms.
(Higher) Expand and simplify \((x+1)(x+2)(x+3)\).
Do two brackets first: \((x^2+3x+2)(x+3)=x^3+3x^2+3x^2+9x+2x+6=x^3+6x^2+11x+6\).
Common error: what is wrong with (x + y)² = x² + y²?
The bracket must be written twice: \((x+y)(x+y)=x^2+2xy+y^2\). The middle term \(2xy\) is the one that gets lost.
Factorise \(12x^2y-18xy^2\) fully.
Take out the highest common factor of the numbers and each letter: \(6xy(2x-3y)\). Expand to check.
Factorise \(x^2-49\) and explain the method.
Difference of two squares: \(a^2-b^2=(a-b)(a+b)\), so \(x^2-49=(x-7)(x+7)\).
Factorise \(x^2+x-12\).
Find two numbers that multiply to \(-12\) and add to \(+1\): \(+4\) and \(-3\). So \((x+4)(x-3)\).
(Higher) Factorise \(2x^2+7x+3\).
Multiply \(2\times3=6\); find two numbers with product 6 and sum 7 (6 and 1). Split: \(2x^2+6x+x+3=2x(x+3)+1(x+3)=(2x+1)(x+3)\).
Solve \(x^2-5x+6=0\) by factorising.
\((x-2)(x-3)=0\), so \(x=2\) or \(x=3\). Always get zero on one side before factorising.
Common error: how many solutions does \(x^2=25\) have?
Two: \(x=5\) and \(x=-5\). Taking only the positive root loses a solution.
(Higher) State the quadratic formula and what each letter means.
For \(ax^2+bx+c=0\): \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\), where \(a\), \(b\), \(c\) are the coefficients (with their signs). Put the whole of \(-b\) and the square root over \(2a\).
(Higher) Solve \(x^2+4x-3=0\) to 2 decimal places.
\(x=\dfrac{-4\pm\sqrt{16+12}}{2}=\dfrac{-4\pm\sqrt{28}}{2}=-2\pm\sqrt7\). So \(x=0.65\) or \(x=-4.65\).
(Higher) Write \(x^2+6x+5\) in the form \((x+p)^2+q\) and state the minimum point.
Half of 6 is 3: \((x+3)^2-9+5=(x+3)^2-4\). The minimum is at \((-3,-4)\).
(Higher) What does the discriminant \(b^2-4ac\) tell you?
If \(b^2-4ac>0\): two distinct real roots. If \(=0\): one repeated root (graph touches the axis). If \(<0\): no real roots (graph does not reach the axis).
Solve \(5x-7=2x+11\).
Collect \(x\) terms on one side: \(3x=18\), so \(x=6\). Do the same operation to both sides.
Solve \(\dfrac{x+3}{2}=\dfrac{2x-1}{3}\).
Cross-multiply: \(3(x+3)=2(2x-1)\), so \(3x+9=4x-2\) and \(x=11\). Check: both sides equal 7.
Solve 3x + 2y = 16 and x − 2y = 0 simultaneously.
Add the equations to eliminate y: 4x = 16, so x = 4. Then 4 − 2y = 0 gives y = 2. Check in the first equation: 12 + 4 = 16.
(Higher) Solve \(y=x+1\) and \(x^2+y^2=25\).
Substitute: \(x^2+(x+1)^2=25\), so \(2x^2+2x-24=0\), \(x^2+x-12=0\), \(x=3\) or \(-4\). The pairs are \((3,4)\) and \((-4,-3)\); give both.
Solve \(3x-4\le11\) and show the solution on a number line.
\(3x\le15\), so \(x\le5\). Use a filled circle at 5 for \(\le\) or \(\ge\), an open circle for \(<\) or \(>\), and shade the line to the left.
What must you do when multiplying or dividing an inequality by a negative number? Solve \(-2x>8\).
Reverse the inequality sign. \(-2x>8\) gives \(x<-4\).
(Higher) Solve the inequality \(x^2-x-6<0\).
\((x-3)(x+2)<0\). The graph is a \(\cup\) shape below the axis between the roots: \(-2
Make \(a\) the subject of \(v^2=u^2+2as\).
Subtract \(u^2\): \(v^2-u^2=2as\). Divide by \(2s\): \(a=\dfrac{v^2-u^2}{2s}\).
(Higher) Make \(x\) the subject of \(y=\dfrac{x+1}{x-2}\).
\(y(x-2)=x+1\), so \(xy-2y=x+1\). Collect: \(xy-x=1+2y\). Factorise: \(x(y-1)=1+2y\). So \(x=\dfrac{1+2y}{y-1}\).
(Higher) Simplify \(\dfrac{x^2-9}{x^2+x-6}\).
Factorise top and bottom: \(\dfrac{(x-3)(x+3)}{(x+3)(x-2)}=\dfrac{x-3}{x-2}\). Only cancel factors, never single terms.
(Higher) Write \(\dfrac1{x+1}+\dfrac2{x-3}\) as a single fraction.
Common denominator \((x+1)(x-3)\): \(\dfrac{(x-3)+2(x+1)}{(x+1)(x-3)}=\dfrac{3x-1}{(x+1)(x-3)}\).
Simplify \((2x^3)^4\).
Raise the number and the power: \(2^4x^{3\times4}=16x^{12}\).
What is the difference between an equation and an identity?
An equation is true only for certain values of \(x\). An identity, written with \(\equiv\), is true for every value, for example \((x+2)^2\equiv x^2+4x+4\).
(Higher) Prove that the sum of any three consecutive integers is a multiple of 3.
Let the integers be \(n\), \(n+1\), \(n+2\). Sum \(=3n+3=3(n+1)\), which is 3 times an integer, so it is a multiple of 3.
Find \(A\) when \(A=\tfrac12(a+b)h\), \(a=5\), \(b=7\), \(h=4\).
\(A=\tfrac12\times12\times4=24\). Substitute carefully and include units if given.
A rectangle has length (x + 3) cm and width x cm. Its perimeter is 26 cm. Find x.
\(2(x+3+x)=26\), so \(4x+6=26\) and \(x=5\). The rectangle is 8 cm by 5 cm.
(Higher) \(y\) is directly proportional to \(x^2\), and \(y=18\) when \(x=3\). Find \(y\) when \(x=5\).
\(y=kx^2\): \(18=9k\), \(k=2\). Then \(y=2\times25=50\).

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