Questions & explanations
1. If the quadratic equation \(x^2 + (k-2)x + (k+1) = 0\) has equal roots, what is the value of \(k\)?
- 0
- 2
- 4
- 8
- 10
Answer: 8
For equal roots, discriminant \(b^2 - 4ac = 0\). Here \(a=1\), \(b=k-2\), \(c=k+1\). So \((k-2)^2 - 4(1)(k+1) = 0\) → \(k^2 -4k +4 -4k -4 = 0\) → \(k^2 -8k = 0\) → \(k(k-8)=0\). Thus \(k=0\) or \(k=8\). But if \(k=0\), the equation becomes \(x^2 -2x +1=0\) which has equal roots \(x=1\). If \(k=8\), equation becomes \(x^2 +6x +9=0\) with equal root \(x=-3\). Both are valid, but only \(k=8\) is among the options.
2. For what values of \(m\) does the quadratic equation \(mx^2 + 2x + 1 = 0\) have two distinct real roots?
- m < 1
- m < 1 and m ≠ 0
- m > 1
- m < 1 and m > 0
- m < 1 and m < 0
Answer: m < 1 and m ≠ 0
For two distinct real roots, discriminant \(b^2 - 4ac > 0\) and \(a \neq 0\). Here \(a=m\), \(b=2\), \(c=1\). Discriminant: \(4 - 4m > 0\) → \(4 > 4m\) → \(m < 1\). Also \(m \neq 0\) because if \(m=0\), the equation becomes linear \(2x+1=0\), not quadratic. So \(m < 1\) and \(m \neq 0\).
3. Solve for x: (x^2 - 3x)^2 - 2(x^2 - 3x) - 8 = 0. Which of the following is the complete set of real solutions?
- {-1, 4}
- {-2, 1, 2, 4}
- {-1, 2, 4}
- {-2, 1, 4}
- {-1, 1, 4}
Answer: {-1, 2, 4}
Let u = x^2 - 3x. Then u^2 - 2u - 8 = 0 → (u-4)(u+2)=0 → u=4 or u=-2. For u=4: x^2-3x-4=0 → (x-4)(x+1)=0 → x=4 or x=-1. For u=-2: x^2-3x+2=0 → (x-1)(x-2)=0 → x=1 or x=2. All four satisfy original equation; no extraneous solutions. Set: {-1, 1, 2, 4}.
4. Solve for \(x\): \(4^x - 2^{x+1} - 8 = 0\)
- x = 2
- x = 3
- x = 1
- x = -1
- x = 0
Answer: x = 2
Rewrite: \(4^x = (2^2)^x = 2^{2x}\), and \(2^{x+1} = 2 \cdot 2^x\). Let \(y = 2^x\), then equation becomes \(y^2 - 2y - 8 = 0\). Factor: \((y-4)(y+2)=0\) → \(y=4\) or \(y=-2\). Since \(y=2^x > 0\), \(y=4\) → \(2^x = 4\) → \(x=2\).
5. If the roots of \(2x^2 - 3x + 1 = 0\) are \(p\) and \(q\), find the quadratic equation whose roots are \(\frac{1}{p}\) and \(\frac{1}{q}\).
- x^2 - 3x + 2 = 0
- x^2 + 3x + 2 = 0
- x^2 - 3x - 2 = 0
- x^2 + 3x - 2 = 0
- 2x^2 - 3x + 1 = 0
Answer: x^2 - 3x + 2 = 0
For \(2x^2 - 3x + 1 = 0\), sum \(p+q = 3/2\), product \(pq = 1/2\). New roots: sum = \(1/p + 1/q = (p+q)/(pq) = (3/2)/(1/2) = 3\). Product = \(1/(pq) = 2\). So equation: \(x^2 - (sum)x + product = 0\) → \(x^2 - 3x + 2 = 0\).
6. A ball is thrown upward from a height of 5 meters with an initial velocity of 20 m/s. Its height h (in meters) after t seconds is given by h = -5t^2 + 20t + 5. The ball hits the ground when h = 0. Determine the time it takes for the ball to hit the ground, and verify that the solution is feasible.
- t = 2 + √5 seconds (approximately 4.24 s)
- t = 2 - √5 seconds (approximately -0.24 s)
- t = 2 + √5 and t = 2 - √5 seconds
- t = 4 seconds
- t = 2 seconds
Answer: t = 2 + √5 seconds (approximately 4.24 s)
Set -5t^2+20t+5=0 → divide by -5: t^2-4t-1=0. Discriminant: 16+4=20 → √20=2√5. Solutions: t = [4 ± 2√5]/2 = 2 ± √5. t = 2+√5 ≈ 4.24 s (positive, feasible), t = 2-√5 ≈ -0.24 s (negative, extraneous). So only t = 2+√5.
7. The roots of the quadratic equation \(x^2 - 3x + k = 0\) are \(\alpha\) and \(\beta\). If \(\alpha^2 + \beta^2 = 7\), what is the value of \(k\)?
- 1
- 2
- 3
- 4
- 5
Answer: 1
For \(x^2 - 3x + k = 0\), sum of roots \(\alpha + \beta = 3\), product \(\alpha\beta = k\). We know \(\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 9 - 2k = 7\). So \(9 - 2k = 7\) → \(2k = 2\) → \(k = 1\).
8. Solve for x: √(x + 5) = x - 1.
- x = 4 only
- x = -1 only
- x = 4 or x = -1
- x = 3 only
- x = 6 only
Answer: x = 4 only
Square both sides: x + 5 = (x - 1)^2 → x + 5 = x^2 - 2x + 1 → 0 = x^2 - 3x - 4 → (x - 4)(x + 1) = 0 → x = 4 or x = -1. Check: x = 4 gives √9 = 3 = 4 - 1, valid. x = -1 gives √4 = 2 ≠ -2, extraneous. Only x = 4.
9. Solve for x: √(x+2) = x. Which of the following is the solution set?
- x = -1
- x = 2
- x = -1 and x = 2
- x = 1
- x = -2
Answer: x = 2
Square both sides: x+2 = x^2 → x^2 - x - 2 = 0 → (x-2)(x+1)=0 → x=2 or x=-1. Check for extraneous: x=2 gives √4=2, works. x=-1 gives √1=1, but -1 ≠ 1, so extraneous. Only x=2 is valid.
10. A rectangular garden has a length that is 4 meters more than its width. The area of the garden is 96 square meters. Is it possible to have a rectangular garden with these dimensions? Use the discriminant to determine the feasibility.
- Yes, because the discriminant is positive and a perfect square.
- Yes, because the discriminant is positive but not a perfect square.
- No, because the discriminant is zero.
- No, because the discriminant is negative.
- Yes, because the discriminant is negative.
Answer: Yes, because the discriminant is positive and a perfect square.
Let width = w, length = w+4. Area: w(w+4)=96 → w^2+4w-96=0. Discriminant: 4^2 - 4(1)(-96)=16+384=400, a positive perfect square (20^2). Thus real rational solutions exist, so feasible.
11. Rewrite the quadratic equation x^2 + 6x + 5 = 0 in vertex form by completing the square. Which of the following is the correct vertex form?
- (x + 3)^2 = 4
- (x + 3)^2 = -4
- (x - 3)^2 = 4
- (x + 3)^2 = 14
- (x - 3)^2 = -4
Answer: (x + 3)^2 = 4
Start with x^2 + 6x + 5 = 0. Move constant: x^2 + 6x = -5. Complete square: add (6/2)^2 = 9 to both sides: x^2 + 6x + 9 = 4. Left side is (x + 3)^2, so vertex form is (x + 3)^2 = 4.
12. Solve for x: 0.2x^2 - 0.5x - 0.3 = 0. Which of the following is a solution?
- -0.5
- 0.5
- 1.5
- 3
- -3
Answer: 3
Multiply equation by 10: 2x^2 - 5x - 3 = 0. Use quadratic formula: x = [5 ± √(25 + 24)]/(4) = [5 ± √49]/4 = [5 ± 7]/4. Solutions: (5+7)/4 = 3, (5-7)/4 = -0.5. Thus 3 is a solution.