Solving equations graphically and analytically practice questions.

Solve the equation 2^x = 8√2.

• x = 4
• x = 3
• x = 3.5
• x = 2.5

Answer: x = 3.5

Write both sides as powers of 2: 8√2 = 2³ × 2^(1/2) = 2^(7/2), so x = 7/2 = 3.5.

Determine the value of k for which the equation x² − 4x + k = 0 has exactly one real solution.

• k = 2
• k = 4
• k = −4
• k = 16

Answer: k = 4

One repeated solution requires the discriminant to be zero: (−4)² − 4k = 0, so 16 = 4k and k = 4.

By considering the graph of y = x³ − 3x, determine how many real solutions the equation x³ − 3x = 1 has.

• 0
• 1
• 2
• 3

Answer: 3

The curve has a local maximum value 2 (at x = −1) and a local minimum value −2 (at x = 1); since −2 < 1 < 2, the horizontal line y = 1 crosses the curve three times.

Solve ln x + ln(x − 2) = ln 8.

• x = 4 only
• x = 4 and x = −2
• x = −2 only
• No real solution

Answer: x = 4 only

Combining logs gives x(x − 2) = 8, so x² − 2x − 8 = 0 and x = 4 or x = −2; but x = −2 makes ln x undefined, so only x = 4 is valid.

Solve the equation 9^x − 4 × 3^x + 3 = 0.

• x = 1 and x = 3
• x = 0 only
• x = 0 and x = 1
• x = ln 3 and x = ln 1

Answer: x = 0 and x = 1

Substituting u = 3^x gives u² − 4u + 3 = 0, so u = 1 or u = 3; then 3^x = 1 gives x = 0 and 3^x = 3 gives x = 1.

The graphs of y = x + 1 and y = 6/x intersect at the points where

• x = −2 and x = 3
• x = 2 and x = −3
• x = 2 only
• x = 1 and x = 6

Answer: x = 2 and x = −3

Setting x + 1 = 6/x and multiplying by x gives x² + x − 6 = 0, which factorises as (x + 3)(x − 2) = 0, so x = 2 or x = −3.