Method 1 - Decimal search

The equation already has 0 on one side; $x^{4} - x^{2} - 9 = 0$ .
Substitute 1.0, 1.1, 1.2, 1.3 ... into the equation until a change of sign happens.

$x$	$x^{4} - x^{2} - 9$
1.0	-9
1.1	-8.7459
1.2	-8.3664
1.3	-7.8339
1.4	-7.1184
1.5	-6.1875
1.6	-5.0064
1.7	-3.5379
1.8	-1.7424
1.9	0.4221

One correct evaluation between 1 and 2 [1]
Two correct evaluations between 1 and 2 [1]

There is a sign change between 1.8 and 1.9. So we now test the midpoint, 1.85

when $x = 1.85, x^{4} - x^{2} - 9 = - 0.708 993 . . .$

[1]

Therefore the sign change is between 1.85 and 1.9 so the solution will round to 1.9 to one decimal place.

$x = 1.9 (1 d . p .)$ [1]

Method 2 - Interval bisection

The equation already has 0 on one side; $x^{4} - x^{2} - 9 = 0$ .
1.5 bisects the interval 1 to 2.
Substitute 1, 1.5 and 2 into the (left-hand side of the) equation.

$x$	$x^{4} - x^{2} - 9$
1	-9
1.5	-6.1875
2	3

First correct evaluation between 1 and 2 [1]

There is a change of sign between 1.5 and 2. Bisect this interval (1.75) and test this value in the equation.

when $x = 1.75, x^{4} - x^{2} - 9 = - 2.683 593 . . .$

Second correct evaluation between 1 and 2 [1]

There is a change of sign between 1.75 and 2. Bisect this interval (1.875) and test this value in the equation.

when $x = 1.875, x^{4} - x^{2} - 9 = - 0.156 005 . . .$

There is a change of sign between 1.875 and 2. Bisect this interval ( (1.875 + 2) ÷ 2 = 1.9375) and test this value in the equation.

when $x = 1.9375, x^{4} - x^{2} - 9 = 1.337 905 . . .$

[1]

There is a change of sign between 1.875 and 1.9375 so the solution will round to 1.9 to one decimal place.

$x = 1.9 (1 d . p .)$ [1]

Iteration (OCR GCSE Maths)

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