This note covers the two ways to solve an equation: analytically, using algebra to find exact answers, and graphically, using your GDC to find the intersection of two graphs. Choosing the right method quickly saves time and avoids lost solutions.
Key ideas
To solve \(f(x)=g(x)\), you are looking for the \(x\)-values where the graphs of \(y=f(x)\) and \(y=g(x)\) meet. To solve \(f(x)=0\), you are looking for where the graph of \(y=f(x)\) crosses the \(x\)-axis. These \(x\)-values are called the roots of the equation, or the zeros of the function.
Graphical method. Enter each side of the equation as a separate function on your GDC. Choose a window that shows all the crossings, then use the intersection feature. If you have rearranged the equation to \(f(x)=0\), use the zero (root) feature instead. Many GDCs also have an equation solver, which is quick but may only return one solution unless you give it different starting values. Always compare the number of solutions with what the graph shows. Give answers to 3 s.f. unless told otherwise.
Analytical method. Choose the technique that fits the equation:
- Linear equations: collect the terms in \(x\) on one side and isolate \(x\).
- Quadratic equations: rearrange to \(ax^2+bx+c=0\), then factorise, complete the square, or use the quadratic formula.
- Higher-degree polynomials: factorise, using a common factor or a factor you have found, or use your GDC polynomial solver.
- Disguised quadratics: if an expression and its square both appear, such as \(4^x=(2^x)^2\), substitute \(u=2^x\), solve the quadratic in \(u\), then go back to \(x\).
- Exponentials and logarithms: take logs, or rewrite in exponential form.
How many solutions? For \(ax^2+bx+c=0\), the discriminant \(\Delta=b^2-4ac\) tells you. If \(\Delta\gt 0\) there are two distinct real roots. If \(\Delta=0\) there is one repeated root, and the graph just touches the \(x\)-axis. If \(\Delta\lt 0\) there are no real roots. This is the standard way to find a parameter value that gives exactly one solution.
Context problems. When an equation comes from a real situation, check each answer against the context. A negative time or length is rejected, and an answer that gives a fraction of a person is usually rounded in a sensible direction.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Root or zero of \(f\) | A value of \(x\) such that \(f(x)=0\). It is an \(x\)-intercept of the graph. |
| \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) | Quadratic formula for \(ax^2+bx+c=0\) with \(a\ne 0\). |
| \(\Delta=b^2-4ac\) | Discriminant. It decides how many real roots a quadratic has. |
| \(ab=0\Rightarrow a=0\) or \(b=0\) | Used after factorising. Never divide both sides by an expression containing \(x\), because you may lose a root. |
| Substitution \(u=a^x\) | Turns \(a^{2x}+pa^x+q=0\) into the quadratic \(u^2+pu+q=0\). Remember \(u\gt 0\). |
| Intersection of \(y=f(x)\) and \(y=g(x)\) | GDC feature for solving \(f(x)=g(x)\) when algebra is not possible. |
Worked examples
Example 1 A basketball is thrown upwards. Its height in metres after \(t\) seconds is \(h=2+14t-4.9t^2\). Find the times when the height is 8 m.
Set \(h=8\): \(2+14t-4.9t^2=8\).
Rearrange to the form \(at^2+bt+c=0\) by moving every term to the right: \(0=4.9t^2-14t+6\).
Here \(a=4.9\), \(b=-14\) and \(c=6\). The discriminant is \(\Delta=(-14)^2-4(4.9)(6)=196-117.6=78.4\), which is positive, so there are two solutions.
\(t=\dfrac{14\pm\sqrt{78.4}}{2(4.9)}=\dfrac{14\pm 8.8544}{9.8}\).
So \(t=\dfrac{5.1456}{9.8}=0.525\) or \(t=\dfrac{22.8544}{9.8}=2.33\).
Both are positive, so both are valid: the ball passes 8 m on the way up and again on the way down.
Answer: \(t=0.525\) s and \(t=2.33\) s
Example 2 Solve \(4^x-5\times 2^x+4=0\).
Notice that \(4^x=(2^2)^x=(2^x)^2\). Let \(u=2^x\).
The equation becomes \(u^2-5u+4=0\).
Factorise: \((u-1)(u-4)=0\), so \(u=1\) or \(u=4\).
Now return to \(x\). If \(2^x=1\), then \(x=0\). If \(2^x=4\), then \(x=2\).
Check \(x=0\): \(1-5+4=0\). Check \(x=2\): \(16-20+4=0\).
Answer: \(x=0\) or \(x=2\)
Example 3 Solve \(e^x=5-x^2\), giving your answers to 3 s.f.
This mixes an exponential with a quadratic, so there is no algebraic method. Use the graph.
Enter \(y=e^x\) as the first function and \(y=5-x^2\) as the second. A window of about \(-4\le x\le 3\) and \(-2\le y\le 8\) shows both curves.
The graph of \(e^x\) rises from near zero, and the downward parabola crosses it twice, once for negative \(x\) and once for positive \(x\).
Use the intersection feature of your GDC at each crossing. The \(x\)-coordinates are \(-2.2114\ldots\) and \(1.2411\ldots\)
Check the second: \(e^{1.2411}=3.460\) and \(5-1.2411^2=3.460\). Check the first: \(e^{-2.2114}=0.1095\) and \(5-2.2114^2=0.1095\).
Answer: \(x=-2.21\) or \(x=1.24\)
Common errors
- Dividing both sides by an expression in \(x\). For \(x^2=3x\), dividing by \(x\) gives only \(x=3\) and loses \(x=0\). Instead write \(x^2-3x=0\) and factorise to \(x(x-3)=0\).
- Forgetting to convert back to \(x\) after a substitution. Finding \(u\) is only half of the problem.
- Keeping an impossible value of \(u\). For \(u=2^x\), a solution such as \(u=-3\) has no real \(x\), because \(2^x\) is always positive.
- Using the quadratic formula before the equation equals zero. Rearrange first so that one side is \(0\).
- Reading only one intersection from the GDC. Check the window and look for every crossing in the domain given.
Exam tip: When a question says "hence" or "use your GDC", writing the equation you entered and quoting the intersection or root you found shows your method, even if the algebra would be impossible.
Check yourself
- Solve \(2x^2+x-6=0\) by factorising.
Answer
\(2x^2+x-6=(2x-3)(x+2)\), because \(2x\times x=2x^2\), \(-3\times 2=-6\) and \(4x-3x=x\). So \(x=\tfrac32\) or \(x=-2\). - Solve \(9^x-4\times 3^x-45=0\).
Answer
Let \(u=3^x\), so \(9^x=u^2\). Then \(u^2-4u-45=0\), which factorises as \((u-9)(u+5)=0\). So \(u=9\) or \(u=-5\). Since \(3^x\gt 0\), reject \(u=-5\). From \(3^x=9\), \(x=2\). - Use your GDC to solve \(\ln x=x-3\), giving answers to 3 s.f.
Answer
Plot \(y=\ln x\) and \(y=x-3\). They cross twice, at \(x=0.0525\) and \(x=4.51\). For the second, \(\ln 4.5052=1.5052\) and \(4.5052-3=1.5052\). - Find the values of \(k\) for which \(x^2+kx+9=0\) has exactly one solution.
Answer
One solution means \(\Delta=0\). Here \(\Delta=k^2-4(1)(9)=k^2-36\). Setting \(k^2-36=0\) gives \(k=6\) or \(k=-6\).