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15 (a) Show that the equation $x^2 + 7x - 5 = 0$ has a solution between $x = 0$ and $x = 1$ - Edexcel - GCSE Maths - Question 15 - 2017 - Paper 3
Question 15
15 (a) Show that the equation $x^2 + 7x - 5 = 0$ has a solution between $x = 0$ and $x = 1$. (b) Show that the equation $x^2 + 7x - 5 = 0$ can be arranged to give $... show full transcript
Worked Solution & Example Answer:15 (a) Show that the equation $x^2 + 7x - 5 = 0$ has a solution between $x = 0$ and $x = 1$ - Edexcel - GCSE Maths - Question 15 - 2017 - Paper 3
Step 1
Show that the equation $x^2 + 7x - 5 = 0$ has a solution between $x = 0$ and $x = 1$
Answer
Let's evaluate the function at the endpoints:
-
At :
[ f(0) = 0^2 + 7(0) - 5 = -5 ] -
At :
[ f(1) = 1^2 + 7(1) - 5 = 3 ]
Since and , there is a sign change between and . By the Intermediate Value Theorem, there exists at least one root in this interval.
Step 2
Step 3
Starting with $x_1 = 1$, use the iteration formula $x_{n} = \frac{5}{x_{n-1} + 7}$ three times to find an estimate for the solution of $x^2 + 7x - 5 = 0$
Answer
Let’s perform the iterations:
- Starting with :
[ x_2 = \frac{5}{1 + 7} = \frac{5}{8} = 0.625 ] - Then, using :
[ x_3 = \frac{5}{0.625 + 7} = \frac{5}{7.625} \approx 0.6563 ] - Finally, using :
[ x_4 = \frac{5}{0.6563 + 7} \approx 0.6590 ]
Thus, after three iterations, .
Step 4
By substituting your answer to part (c) into $x^2 + 7x - 5$, comment on the accuracy of your estimate for the solution to $x^2 + 7x - 5 = 0$
Answer
Substituting into the original equation:
[ (0.6590)^2 + 7(0.6590) - 5 \approx 0.4343 + 4.6130 - 5 \approx 0.0473 ]
This result is close to zero, indicating that our estimate is reasonably accurate, but slightly off. The deviation suggests the root lies very close to our approximation.