WAEC 2025 · Paper 2 · Q7✱✱

(a)
Given $p = x + ym^3$ , find $m$ in terms of $p$ , $x$ and $y$ .
Show the answer
$m = \sqrt[3]{\dfrac{p - x}{y}}$
(b)
By completing the square, find the roots of $x^2 - 6x + 7 = 0$ , correct to one decimal place.
Separate values with commas, e.g. 3, −2
(c)
The product of two consecutive positive odd numbers is 195. Form a quadratic equation and solve it to find the numbers.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Take $x$ from both sides: $ym^3 = p - x$ .
Divide by $y$ : $m^3 = \frac{p - x}{y}$ .
Take the cube root: $m = \sqrt[3]{\frac{p - x}{y}}$ .

(b)

Move the number across: $x^2 - 6x = -7$ .
Half of $-6$ is $-3$ , and $(-3)^2 = 9$ : add 9 to both sides.
$x^2 - 6x + 9 = 2$ , so $(x - 3)^2 = 2$ .
Take the square root, remembering $\pm$ : $x - 3 = \pm\sqrt2$ .
So $x = 3 + \sqrt2 \approx 4.4$ or $x = 3 - \sqrt2 \approx 1.6$ (to one decimal place).

(c)

Let the smaller odd number be $n$ .
The next odd number is $n + 2$ .
Their product is 195: $n(n + 2) = 195$ , so $n^2 + 2n - 195 = 0$ .
Two numbers that multiply to $-195$ and add to 2 are 15 and $-13$ : $(n + 15)(n - 13) = 0$ .
So $n = 13$ or $n = -15$ .
The numbers are positive, so $n = 13$ : the numbers are 13 and 15.

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