Simultaneous equations, inequalities and polynomials – Year 12 A level maths revision notes to download as PDF

Worked example 1

Solve \(3x+2y=16\) and \(x-y=2\).

Write \(x=y+2\), so \(3(y+2)+2y=16\), \(5y=10\), \(y=2\) and \(x=4\).

Worked example 2

Solve \(y=x+1\) and \(x^2+y^2=25\).

Substitute: \(x^2+(x+1)^2=25\), so \(2x^2+2x-24=0\), \(x^2+x-12=0\), \((x+4)(x-3)=0\). Solutions: \(x=3,y=4\) and \(x=-4,y=-3\).

Worked example 3

Find \(k\) if \(y=3x+k\) is a tangent to \(y=x^2+2x+5\).

Setting equal: \(x^2-x+(5-k)=0\). Tangent means \(b^2-4ac=0\): \(1-4(5-k)=0\), so \(k=\dfrac{19}{4}\).

Worked example 4

Solve \(x^2-x-6<0\).

\((x-3)(x+2)<0\), roots \(-2\) and \(3\). The \(\cup\)-shaped curve is below the axis between the roots, so \(-2\lt x<3\).

Worked example 5

Solve \(x^2+3x-10\ge0\).

The roots are \(-5\) and \(2\) and the curve is above the axis outside them: \(x\le-5\) or \(x\ge2\). In set notation \(\{x:x\le-5\}\cup\{x:x\ge2\}\).

Worked example 6

Find the values of \(x\) with \(x^2-5x+4<0\) and \(3x-2>1\).

First: \((x-1)(x-4)<0\), so \(1\lt x<4\). Second: \(3x>3\), so \(x>1\). Both must hold, so the answer is \(1\lt x<4\).

Worked example 7

Factorise \(f(x)=x^3-3x^2-4x+12\).

\(f(2)=8-12-8+12=0\), so \((x-2)\) is a factor. Write \(x^3-3x^2-4x+12=(x-2)(x^2+px+q)\): \(p-2=-3\) gives \(p=-1\), and \(-2q=12\) gives \(q=-6\). Check the \(x\) term: \(q-2p=-4\). So \(f(x)=(x-2)(x-3)(x+2)\).

Worked example 8: Simultaneous equations by elimination

Solve \(3x+2y=16\) and \(5x-4y=6\).

Multiply the first equation by 2 so that the \(y\) coefficients match in size: \(6x+4y=32\). Add this to the second equation: \(6x+4y+5x-4y=32+6\), so \(11x=38\) and \(x=\tfrac{38}{11}\). Substitute back: \(3\times\tfrac{38}{11}+2y=16\) gives \(2y=\tfrac{176-114}{11}=\tfrac{62}{11}\), so \(y=\tfrac{31}{11}\). Always check in the other equation: \(5\times\tfrac{38}{11}-4\times\tfrac{31}{11}=\tfrac{190-124}{11}=6\) ✓.

Worked example 9: A line meeting a curve at two points

Find where the line \(y=x+1\) meets the curve \(y=x^2-3x+5\) and the length of the chord between the points.

Equate: \(x^2-3x+5=x+1\), so \(x^2-4x+4=0\), \((x-2)^2=0\). There is one repeated solution \(x=2\), \(y=3\): the line is a tangent at \((2,3)\) and there is no chord. Change the line to \(y=x-1\): \(x^2-4x+6=0\) has discriminant \(16-24<0\), so there are no intersections. Try \(y=x+3\): \(x^2-4x+2=0\), \(x=2\pm\sqrt2\), giving two points whose distance is \(\sqrt{2\times(2\sqrt2)^2}=4\).

Worked example 10: Algebraic long division by a linear factor

Divide \(2x^3-5x^2-4x+3\) by \((x-3)\) and use the result to solve \(2x^3-5x^2-4x+3=0\).

\(2x^3\div x=2x^2\); subtract \(2x^3-6x^2\) to leave \(x^2-4x\). \(x^2\div x=x\); subtract \(x^2-3x\) to leave \(-x+3\). \(-x\div x=-1\); subtract \(-x+3\) to leave 0. So the quotient is \(2x^2+x-1\) with no remainder, and \(2x^3-5x^2-4x+3=(x-3)(2x^2+x-1)=(x-3)(2x-1)(x+1)\). The roots are \(x=3\), \(x=\tfrac12\) and \(x=-1\).

Worked example 11: Dividing by a non-monic linear factor

Find the remainder when \(6x^3+x^2-5x+4\) is divided by \((2x-1)\).

By the remainder theorem, divide by writing \(2x-1=2(x-\tfrac12)\) and evaluate at \(x=\tfrac12\). \(p(\tfrac12)=6\times\tfrac18+\tfrac14-\tfrac52+4=\tfrac34+\tfrac14-\tfrac52+4=2.5\). So the remainder is \(\tfrac52\). Check by division: \(6x^3+x^2-5x+4=(2x-1)(3x^2+2x-\tfrac32)+\tfrac52\). The key idea is that the remainder on division by \((ax-b)\) is \(p\left(\tfrac ba\right)\).

Worked example 12: A cubic inequality

Solve \((x+2)(x-1)(x-4)\ge0\).

The critical values are \(-2\), \(1\) and \(4\). The cubic has a positive \(x^3\) coefficient, so it is positive for \(x>4\), and the sign alternates as you move left past each root: negative for \(1\lt x<4\), positive for \(-2\lt x<1\), negative for \(x<-2\). The inequality allows equality, so the roots are included: \(-2\le x\le1\) or \(x\ge4\). Test \(x=0\): \(2\times(-1)\times(-4)=8>0\) ✓, test \(x=2\): \(4\times1\times(-2)=-8<0\) ✓ (so excluded).

Worked example 13: Describing a region

Describe the region defined by \(y>x^2-4\) and \(y\le x+2\).

Draw the parabola \(y=x^2-4\) as a dashed curve (strict inequality, boundary not included) and the line \(y=x+2\) as a solid line (boundary included). The line meets the parabola where \(x^2-4=x+2\), i.e. \(x^2-x-6=0\), \((x-3)(x+2)=0\), so \(x=3\) and \(x=-2\), giving points \((3,5)\) and \((-2,0)\). The region lies above the parabola and on or below the line: it is the finite lens-shaped area between \(x=-2\) and \(x=3\). Test the origin: \(0>-4\) ✓ and \(0\le2\) ✓, so the origin is inside the region.

Worked example 14: Finding the equation of a cubic from its roots

A cubic curve crosses the \(x\)-axis at \(x=-1\), \(x=2\) and \(x=5\) and passes through the point \((0,20)\). Find its equation and the height of the curve at \(x=3\).

Write the cubic in factored form \(y=k(x+1)(x-2)(x-5)\). At \(x=0\): \(k(1)(-2)(-5)=10k=20\), so \(k=2\). The equation is \(y=2(x+1)(x-2)(x-5)\), which expands to \(y=2x^3-12x^2+6x+20\). At \(x=3\): \(y=2\times4\times1\times(-2)=-16\), so the curve is 16 units below the axis there, which agrees with the sign pattern: the curve is negative between the roots 2 and 5.

Worked example 15: A repeated factor and a tangent to the axis

Given that \((x-3)^2\) is a factor of \(x^3-5x^2+3x+9\), find the remaining factor and sketch the curve.

Since \((x-3)^2=x^2-6x+9\) is a factor, divide: \(x^3-5x^2+3x+9=(x^2-6x+9)(x+1)\). Check by expanding: \(x^3-6x^2+9x+x^2-6x+9=x^3-5x^2+3x+9\) ✓. So \(y=(x-3)^2(x+1)\). The curve crosses the axis at \(x=-1\) and touches it at \(x=3\) (repeated root), with \(y\)-intercept 9. For \(x>-1\) the curve is never negative because \((x-3)^2\ge0\), so the curve rises from \((-1,0)\), turns at a maximum, comes down to touch the axis at \((3,0)\) and then rises again.

Worked example 16: An inequality in a modelling context

A rectangle has width \(x\) cm and length \((2x+5)\) cm. The area must be greater than 75 cm\(^2\) but the perimeter must not exceed 60 cm. Find the possible values of \(x\).

Area: \(x(2x+5)>75\) gives \(2x^2+5x-75>0\). Use the formula: \(x=\dfrac{-5\pm\sqrt{25+600}}4=\dfrac{-5\pm25}4\), so \(x=5\) or \(x=-7.5\). The U-shape is positive outside the roots; since \(x>0\) we need \(x>5\). Perimeter: \(2(x+2x+5)=6x+10\le60\) gives \(x\le\tfrac{25}3\approx8.33\). Both conditions: \(5\lt x\le8\tfrac13\). For instance \(x=6\) gives a rectangle 6 cm by 17 cm with area 102 cm\(^2\) and perimeter 46 cm.