Quadratic equations and functions — Topic practice

5 exercises · 3 automatically checked · 2 written self-checks · Untimed

Instructions and review guidance

3 questions · Shuffled order · Untimed practice

Work through the set before checking solutions. Guide question labels match the source lesson references. 2 written responses appear below the question bank, with model answers for self-review; these are not automatically graded.

Guide question labels stay the same when the order changes. Check answers when you are ready to review.

Written self-checks are below the question bank and are excluded from the automatic score.

Written responses are self-review exercises.

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The drill

Question 2

  • Guide question E3.2

Find the minimum value of 3(x − 2)2 − 7.

Enter an integer, decimal or fraction without units. Use up to 5 characters, or 6 including a leading minus.

Question 3

  • Guide question E3.3

For what value of c does x2 + 6x + c = 0 have exactly one distinct real solution?

Enter an integer, decimal or fraction without units. Use up to 5 characters, or 6 including a leading minus.

Question 4

  • Guide question E3.4

If r and s are the roots of 2x2 − 7x + 3 = 0, find r2 + s2.

Enter an integer, decimal or fraction without units. Use up to 5 characters, or 6 including a leading minus.

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Written self-checks

Write your complete response before opening the model answer. Compare both your answer and your reasoning.

Guide question E3.1

Solve x2 − 9x + 20 = 0.

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4 and 5

The factors must multiply to 20 and add to −9:

x2 − 9x + 20 = (x − 4)(x − 5) = 0.

The zero-product property gives x = 4 or x = 5. Both satisfy the original equation. Do not reverse the root signs after finding the factors. Review 3.01.

Guide question E3.5

Find all values of k for which kx2 + 2x + 1 = 0 has exactly one distinct real solution.

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0 and 1

Check the degree-changing case first. At k = 0, the equation is 2x + 1 = 0, which has one solution. For k ≠ 0, use the discriminant: D = 22 − 4k = 0 k = 1.

At k = 1, the equation is (x + 1)2 = 0. Both k = 0 and k = 1 qualify. A discriminant-only solution misses the linear case. Review 3.15.