Quadratics

1. Write x^2 - 10x + 7 in completed-square form.

Answer A: (x - 5)^2 - 18

x^2 - 10x + 7 = (x - 5)^2 - 25 + 7 = (x - 5)^2 - 18. Expanding the selected form returns x^2 - 10x + 7.

2. Write 2x^2 + 12x - 1 in completed-square form.

Answer A: 2(x + 3)^2 - 19

2x^2 + 12x - 1 = 2(x^2 + 6x) - 1 = 2[(x + 3)^2 - 9] - 1 = 2(x + 3)^2 - 19.

3. For real x, state the vertex and range of y = -3x^2 + 12x - 5.

Answer A: Vertex (2, 7), range y <= 7

Completing the square gives y = -3(x - 2)^2 + 7. The square term is non-positive after multiplication by -3, so the maximum is 7 at x = 2; hence the vertex is (2, 7) and y <= 7.

4. For x >= 1, find the range of f(x) = (x + 2)^2 - 5.

Answer A: f(x) >= 4

On x >= 1, x + 2 is positive and increases, so f is increasing. Its least permitted value is f(1) = 9 - 5 = 4, giving the range f(x) >= 4.

5. How many real roots does 4x^2 + 4x + 5 = 0 have?

Answer A: No real roots

D = 4^2 - 4(4)(5) = 16 - 80 = -64. Since D < 0, the equation has no real roots.

6. Find k if x^2 + kx + 9 = 0 has a repeated real root.

Answer A: k = 6 or k = -6

Set D = k^2 - 4(1)(9) = k^2 - 36 equal to zero. Then k^2 = 36, so k = 6 or k = -6.

7. For which values of m does x^2 + 2x + m = 0 have two distinct real roots?

Answer B: m < 1

Two distinct real roots require 4 - 4m > 0. Dividing by 4 gives 1 - m > 0, so m < 1. Equality would produce a repeated root.

8. Solve 3x^2 - 11x - 4 = 0.

Answer B: x = 4 or x = -1/3

3x^2 - 11x - 4 = (3x + 1)(x - 4). The zero-product rule gives x = -1/3 or x = 4.

9. Solve x^2 + 6x - 2 = 0 exactly.

Answer B: x = -3 +/- sqrt(11)

x^2 + 6x - 2 = 0 becomes (x + 3)^2 - 11 = 0. Therefore x + 3 = +/-sqrt(11), giving x = -3 +/- sqrt(11).

10. Solve 2x^2 + x - 7 = 0 exactly.

Answer B: x = (-1 +/- sqrt(57))/4

The discriminant is 1^2 - 4(2)(-7) = 57. Hence x = [-1 +/- sqrt(57)]/(2 times 2) = (-1 +/- sqrt(57))/4.

11. Solve 7x^2 - 5x - 4 = 0, giving both roots to 3 significant figures.

Answer B: x = 1.19 or x = -0.479

The exact roots are [5 +/- sqrt(137)]/14. They are approximately 1.19319 and -0.478907, which round to x = 1.19 or x = -0.479 to 3 significant figures.

12. Solve x^2 - 7x + 10 > 0.

Answer B: x < 2 or x > 5

The expression is (x - 2)(x - 5), with roots 2 and 5. It is positive outside the roots, and the strict inequality excludes both endpoints. Thus x < 2 or x > 5.

13. Solve 6 - x - x^2 >= 0.

Answer C: -3 <= x <= 2

The boundary roots are -3 and 2. The downward-opening quadratic is non-negative between them, including the roots because equality is allowed. Therefore -3 <= x <= 2.

14. Solve (x - 1)(x + 4) < 6.

Answer C: -5 < x < 2

Expanding and rearranging gives x^2 + 3x - 10 < 0, or (x + 5)(x - 2) < 0. The upward-opening quadratic is negative between its roots, so -5 < x < 2.

15. Solve y = 3x - 2 and y = x^2 - x + 2 simultaneously.

Answer C: (2, 4) only

Equating gives 3x - 2 = x^2 - x + 2, so x^2 - 4x + 4 = 0 = (x - 2)^2. The repeated root x = 2 gives y = 4, so the only ordered-pair solution is (2, 4).

16. Solve x + y = 5 and xy = 6 simultaneously.

Answer C: (2, 3) and (3, 2)

Substitution gives x(5 - x) = 6, so x^2 - 5x + 6 = 0. Thus x = 2 or 3, giving y = 3 or 2 respectively. Both ordered pairs satisfy both original equations.

17. How many real solutions does the simultaneous system y = x + 4 and y = x^2 + 2 have?

Answer C: Two

Equating gives x^2 - x - 2 = 0, whose discriminant is 9 > 0. Therefore there are two distinct real x-values and hence two real ordered-pair solutions.

18. Solve x^4 - 10x^2 + 9 = 0 for real x.

Answer C: x = -3, -1, 1 or 3

With u = x^2, (u - 1)(u - 9) = 0. Thus x^2 = 1 or x^2 = 9, giving x = -1, 1, -3 or 3.

19. Solve x^(2/3) - 5x^(1/3) + 6 = 0 for real x.

Answer D: x = 8 or x = 27

The substituted equation is u^2 - 5u + 6 = 0, so u = 2 or 3. Reversing u = x^(1/3) gives x = 2^3 = 8 or x = 3^3 = 27.

20. Solve x - 7sqrt(x) + 12 = 0 for real x.

Answer D: x = 9 or x = 16

With u = sqrt(x) >= 0, (u - 3)(u - 4) = 0. Thus u = 3 or 4, so x = 9 or 16. Both values satisfy the original equation.

21. Solve sqrt(x + 6) = x for real x.

Answer D: x = 3 only

The domain requires x >= 0. Squaring gives x + 6 = x^2, so (x - 3)(x + 2) = 0. The candidate -2 violates x >= 0 and is extraneous; x = 3 checks because sqrt(9) = 3.

22. For x != 0, solve 1/x^2 - 5/x + 6 = 0.

Answer D: x = 1/2 or x = 1/3

The equation becomes u^2 - 5u + 6 = 0, so u = 2 or 3. Since u = 1/x, x = 1/2 or 1/3. Both are allowed and satisfy the original equation; x = 0 remains excluded.

23. The equation x^2 + (p + 1)x + p = 0 has a repeated real root. What is p?

Answer D: p = 1

The discriminant is p^2 + 2p + 1 - 4p = (p - 1)^2. A repeated root requires this to be zero, so p = 1.

24. The line y = 2x + k touches the parabola y = x^2 - 4x + 7 at exactly one point. Find k.

Answer D: k = -2

Equating gives x^2 - 6x + (7 - k) = 0. Tangency requires D = 36 - 4(7 - k) = 8 + 4k = 0, so k = -2. The repeated root is x = 3.