Math 2 Quiz: Quadratic Formula And Discriminant
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Quadratic Formula And DiscriminantQuestion 1 of 11

A quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 has discriminant Δ=25\Delta = 25. If the equation is modified to (2a)x2+(2b)x+(2c)=0(2a)x^2 + (2b)x + (2c) = 0, what is the discriminant of the new equation?

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Math 2 Quiz

Math 2 Quiz: Quadratic Formula And Discriminant

Practice Quadratic Formula And Discriminant in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Quadratic Formula And Discriminant, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 has discriminant Δ=25\Delta = 25. If the equation is modified to (2a)x2+(2b)x+(2c)=0(2a)x^2 + (2b)x + (2c) = 0, what is the discriminant of the new equation?

  1. 25
  2. 50
  3. 100 (correct answer)
  4. 200
Explanation: The discriminant of ax2+bx+c=0ax^2 + bx + c = 0 is b2−4ac=25b^2 - 4ac = 25. For the modified equation (2a)x2+(2b)x+(2c)=0(2a)x^2 + (2b)x + (2c) = 0, the discriminant is (2b)2−4(2a)(2c)=4b2−16ac=4(b2−4ac)=4(25)=100(2b)^2 - 4(2a)(2c) = 4b^2 - 16ac = 4(b^2 - 4ac) = 4(25) = 100. Choice A incorrectly assumes the discriminant remains unchanged. Choice B results from the error 2(b2−4ac)2(b^2 - 4ac), incorrectly factoring out only one factor of 2. Choice D results from the error (2b)2−8ac=4b2−8ac(2b)^2 - 8ac = 4b^2 - 8ac, miscalculating the constant term coefficient.

Question 2

The quadratic equation px2+qx+r=0px^2 + qx + r = 0 has discriminant Δ1=36\Delta_1 = 36. If each coefficient is doubled to form 2px2+2qx+2r=02px^2 + 2qx + 2r = 0, and then the equation is rewritten in standard form ax2+bx+c=0ax^2 + bx + c = 0 by dividing by the leading coefficient, what is the discriminant Δ2\Delta_2 of the resulting equation?

  1. Δ2=9\Delta_2 = 9
  2. Δ2=18\Delta_2 = 18
  3. Δ2=36\Delta_2 = 36 (correct answer)
  4. Δ2=72\Delta_2 = 72
Explanation: Starting with px2+qx+r=0px^2 + qx + r = 0 where Δ1=q2−4pr=36\Delta_1 = q^2 - 4pr = 36. After doubling coefficients: 2px2+2qx+2r=02px^2 + 2qx + 2r = 0. Dividing by the leading coefficient 2p2p to get standard form: x2+2q2px+2r2p=0x^2 + \frac{2q}{2p}x + \frac{2r}{2p} = 0, which simplifies to x2+qpx+rp=0x^2 + \frac{q}{p}x + \frac{r}{p} = 0. For this equation x2+qpx+rp=0x^2 + \frac{q}{p}x + \frac{r}{p} = 0, the discriminant is Δ2=(qp)2−4(1)(rp)=q2p2−4rp=q2−4prp2=Δ1p2=36p2\Delta_2 = \left(\frac{q}{p}\right)^2 - 4(1)\left(\frac{r}{p}\right) = \frac{q^2}{p^2} - \frac{4r}{p} = \frac{q^2 - 4pr}{p^2} = \frac{\Delta_1}{p^2} = \frac{36}{p^2}. However, this depends on pp, which seems problematic. Let me reconsider the problem. Actually, I think the key insight is that when we divide 2px2+2qx+2r=02px^2 + 2qx + 2r = 0 by 2p2p, we get x2+qpx+rp=0x^2 + \frac{q}{p}x + \frac{r}{p} = 0. The discriminant of this is Δ2=(qp)2−4(rp)=q2p2−4rp=q2−4prp2=36p2\Delta_2 = \left(\frac{q}{p}\right)^2 - 4\left(\frac{r}{p}\right) = \frac{q^2}{p^2} - \frac{4r}{p} = \frac{q^2 - 4pr}{p^2} = \frac{36}{p^2}. But wait, this still depends on pp. Let me think about this differently. Maybe the problem is asking about the discriminant after scaling. When we have 2px2+2qx+2r=02px^2 + 2qx + 2r = 0, the discriminant is (2q)2−4(2p)(2r)=4q2−16pr=4(q2−4pr)=4×36=144(2q)^2 - 4(2p)(2r) = 4q^2 - 16pr = 4(q^2 - 4pr) = 4 \times 36 = 144. But then when we divide by 2p2p, we need to be more careful. Actually, let me approach this more systematically. The roots of px2+qx+r=0px^2 + qx + r = 0 are the same as the roots of 2px2+2qx+2r=02px^2 + 2qx + 2r = 0 (since we just multiplied by 2). When we divide by 2p2p, we get x2+qpx+rp=0x^2 + \frac{q}{p}x + \frac{r}{p} = 0, which still has the same roots as px2+qx+r=0px^2 + qx + r = 0. The key insight is that dividing the original equation px2+qx+r=0px^2 + qx + r = 0 by pp gives x2+qpx+rp=0x^2 + \frac{q}{p}x + \frac{r}{p} = 0, and this is exactly what we get after the doubling and re-normalizing process. Therefore, Δ2=Δ1=36\Delta_2 = \Delta_1 = 36. The discriminant is preserved because the roots are preserved through these transformations. Option A (9) might result from incorrectly thinking the discriminant scales as 1/41/4 of the original. Option B (18) might come from thinking it scales as 1/21/2. Option D (72) might come from thinking it doubles.

Question 3

If the discriminant of px2+qx+r=0px^2 + qx + r = 0 is Δ\Delta, what is the discriminant of rx2+qx+p=0rx^2 + qx + p = 0?

  1. Δ\Delta (correct answer)
  2. −Δ-\Delta
  3. Δpr\frac{\Delta}{pr}
  4. Δ⋅rp\Delta \cdot \frac{r}{p}
Explanation: The discriminant of px2+qx+r=0px^2 + qx + r = 0 is Δ=q2−4pr\Delta = q^2 - 4pr. The discriminant of rx2+qx+p=0rx^2 + qx + p = 0 is q2−4rp=q2−4pr=Δq^2 - 4rp = q^2 - 4pr = \Delta. The discriminant is unchanged when we swap the coefficients of x2x^2 and the constant term. Choice B suggests a sign change that doesn't occur. Choice C would result from incorrectly dividing by the product prpr. Choice D comes from multiplying by the ratio rp\frac{r}{p}, which might seem intuitive but is incorrect.

Question 4

The quadratic ax2+bx+c=0ax^2 + bx + c = 0 has two roots. If both roots are increased by 3, the new quadratic equation is x2−8x+15=0x^2 - 8x + 15 = 0. What was the sum of the original roots?

  1. −2-2
  2. 1414
  3. 88
  4. 22 (correct answer)
Explanation: This question tests your understanding of how transformations affect polynomial roots and the relationship between roots and coefficients. When you increase both roots of a quadratic by the same amount, you're performing a horizontal shift on the original equation. If the original roots were r1r_1 and r2r_2, the new roots are r1+3r_1 + 3 and r2+3r_2 + 3. For the new quadratic x2−8x+15=0x^2 - 8x + 15 = 0, you can find its roots by factoring: (x−3)(x−5)=0(x - 3)(x - 5) = 0, so the new roots are 3 and 5. This means the original roots were 3−3=03 - 3 = 0 and 5−3=25 - 3 = 2. Therefore, the sum of the original roots is 0+2=20 + 2 = 2. Alternatively, you can use Vieta's formulas. For any quadratic px2+qx+r=0px^2 + qx + r = 0, the sum of roots equals −qp-\frac{q}{p}. In the new equation, this sum is −(−8)1=8-\frac{(-8)}{1} = 8. Since each original root was increased by 3, the sum increased by 2×3=62 \times 3 = 6. So the original sum was 8−6=28 - 6 = 2. Choice A) −2-2 incorrectly subtracts 6 from 2 instead of from 8. Choice B) 1414 mistakenly adds 6 to 8 instead of subtracting. Choice C) 88 is the sum of the new roots, not the original ones. Study tip: When roots are shifted by a constant, remember that the sum of roots changes by (number of roots) × (shift amount). This relationship appears frequently in polynomial transformation problems.

Question 5

A quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c has discriminant 49. If g(x)=f(x+2)g(x) = f(x + 2), what is the discriminant of the equation g(x)=0g(x) = 0?

  1. 49 (correct answer)
  2. 53
  3. 147
  4. 196
Explanation: If f(x)=ax2+bx+cf(x) = ax^2 + bx + c, then g(x)=f(x+2)=a(x+2)2+b(x+2)+c=ax2+(4a+b)x+(4a+2b+c)g(x) = f(x + 2) = a(x + 2)^2 + b(x + 2) + c = ax^2 + (4a + b)x + (4a + 2b + c). The discriminant of g(x)=0g(x) = 0 is (4a+b)2−4a(4a+2b+c)=16a2+8ab+b2−16a2−8ab−4ac=b2−4ac=49(4a + b)^2 - 4a(4a + 2b + c) = 16a^2 + 8ab + b^2 - 16a^2 - 8ab - 4ac = b^2 - 4ac = 49. The discriminant remains unchanged under horizontal translations. Choice B might result from incorrectly adding 22=42^2 = 4 to the discriminant. Choice C comes from multiplying by 3 (perhaps 72÷49×37^2 \div 49 \times 3). Choice D equals (49)2/25(49)^2/25, possibly from squaring and scaling errors.

Question 6

For the quadratic equation x2−2mx+(m2−4)=0x^2 - 2mx + (m^2 - 4) = 0, what condition on mm ensures that both roots are positive?

  1. m>2m > 2 (correct answer)
  2. m≥2m \geq 2
  3. m>0m > 0
  4. m≥0m \geq 0
Explanation: For both roots to be positive, we need: (1) discriminant ≥0\geq 0, (2) sum of roots >0> 0, and (3) product of roots >0> 0. The discriminant is (2m)2−4(m2−4)=4m2−4m2+16=16>0(2m)^2 - 4(m^2 - 4) = 4m^2 - 4m^2 + 16 = 16 > 0 (always satisfied). The sum of roots is 2m2m (from Vieta's formulas), so we need 2m>02m > 0, giving m>0m > 0. The product of roots is m2−4m^2 - 4, so we need m2−4>0m^2 - 4 > 0, giving m2>4m^2 > 4, so m>2m > 2 or m<−2m < -2. Combining m>0m > 0 and (m>2m > 2 or m<−2m < -2), we get m>2m > 2. Choice B includes m=2m = 2, but then the product equals 0, making one root zero. Choice C only ensures positive sum but allows negative product. Choice D includes m=0m = 0, which gives a negative product.

Question 7

For which value of mm does the quadratic equation x2+mx+(m−3)=0x^2 + mx + (m-3) = 0 have roots whose sum of squares is minimized?

  1. m=0m = 0
  2. m=2m = 2
  3. m=1m = 1 (correct answer)
  4. m=3m = 3
Explanation: This problem combines quadratic equations with optimization, requiring you to use Vieta's formulas and express the sum of squares in terms of the parameter mm. For a quadratic x2+mx+(m−3)=0x^2 + mx + (m-3) = 0 with roots rr and ss, Vieta's formulas give us r+s=−mr + s = -m and rs=m−3rs = m-3. To find r2+s2r^2 + s^2, use the algebraic identity: r2+s2=(r+s)2−2rsr^2 + s^2 = (r+s)^2 - 2rs. Substituting our values: r2+s2=(−m)2−2(m−3)=m2−2m+6r^2 + s^2 = (-m)^2 - 2(m-3) = m^2 - 2m + 6. To minimize this expression, take the derivative and set it equal to zero: ddm(m2−2m+6)=2m−2=0\frac{d}{dm}(m^2 - 2m + 6) = 2m - 2 = 0, which gives m=1m = 1. Since the second derivative is positive, this is indeed a minimum. Let's check why the other options fail. Option A (m=0m = 0) gives r2+s2=0−0+6=6r^2 + s^2 = 0 - 0 + 6 = 6. Option B (m=2m = 2) gives r2+s2=4−4+6=6r^2 + s^2 = 4 - 4 + 6 = 6. Option D (m=3m = 3) gives r2+s2=9−6+6=9r^2 + s^2 = 9 - 6 + 6 = 9. With m=1m = 1, we get r2+s2=1−2+6=5r^2 + s^2 = 1 - 2 + 6 = 5, which is indeed the smallest value. When you see optimization problems involving quadratic roots, remember to use Vieta's formulas to express everything in terms of the parameter, then apply standard calculus techniques to find extrema.

Question 8

The equation kx2−12x+k=0kx^2 - 12x + k = 0 has two distinct real roots. Which of the following represents all possible values of kk?

  1. k>6k > 6 or k<−6k < -6
  2. −6<k<6-6 < k < 6 and k≠0k \neq 0 (correct answer)
  3. k>0k > 0 and k≠6k \neq 6
  4. k<0k < 0 or 0<k<60 < k < 6
Explanation: For two distinct real roots, we need k≠0k \neq 0 (to ensure it's quadratic) and discriminant >0> 0. The discriminant is 144−4k2>0144 - 4k^2 > 0, which gives 144>4k2144 > 4k^2, so 36>k236 > k^2, meaning −6<k<6-6 < k < 6. Combined with k≠0k \neq 0, we get −6<k<6-6 < k < 6 and k≠0k \neq 0. Choice A gives the condition for no real roots (discriminant <0< 0). Choice C omits negative values of kk that work. Choice D incorrectly includes k<−6k < -6, which would make the discriminant negative.

Question 9

If α\alpha and β\beta are roots of x2−5x+3=0x^2 - 5x + 3 = 0, what quadratic equation has roots α2\alpha^2 and β2\beta^2?

  1. x2−25x+9=0x^2 - 25x + 9 = 0
  2. x2−19x+81=0x^2 - 19x + 81 = 0
  3. x2−31x+9=0x^2 - 31x + 9 = 0
  4. x2−19x+9=0x^2 - 19x + 9 = 0 (correct answer)
Explanation: When you encounter a problem asking for a quadratic with transformed roots, you need to use Vieta's formulas to find the sum and product of the new roots without actually solving for the original roots. From the given equation x2−5x+3=0x^2 - 5x + 3 = 0, Vieta's formulas tell us that α+β=5\alpha + \beta = 5 and αβ=3\alpha\beta = 3. To find the quadratic with roots α2\alpha^2 and β2\beta^2, you need α2+β2\alpha^2 + \beta^2 and α2β2\alpha^2\beta^2. For the product: α2β2=(αβ)2=32=9\alpha^2\beta^2 = (\alpha\beta)^2 = 3^2 = 9 For the sum: α2+β2=(α+β)2−2αβ=52−2(3)=25−6=19\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 5^2 - 2(3) = 25 - 6 = 19 Therefore, the quadratic is x2−19x+9=0x^2 - 19x + 9 = 0. Choice A (x2−25x+9=0x^2 - 25x + 9 = 0) uses (α+β)2=25(\alpha + \beta)^2 = 25 directly without subtracting 2αβ2\alpha\beta. Choice B (x2−19x+81=0x^2 - 19x + 81 = 0) correctly finds the sum but incorrectly uses (αβ)2(\alpha\beta)^2 twice, giving 34=813^4 = 81 instead of 32=93^2 = 9. Choice C (x2−31x+9=0x^2 - 31x + 9 = 0) appears to add an extra term to the sum calculation. Remember this pattern: when transforming roots by squaring, the new product is the square of the original product, but finding the new sum requires the algebraic identity α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta. This identity is crucial for root transformation problems.

Question 10

If rr and ss are the roots of 2x2−7x+3=02x^2 - 7x + 3 = 0, what is the value of 1r+1s\frac{1}{r} + \frac{1}{s}?

  1. 37\frac{3}{7}
  2. 73\frac{7}{3} (correct answer)
  3. 76\frac{7}{6}
  4. 32\frac{3}{2}
Explanation: Using Vieta's formulas for 2x2−7x+3=02x^2 - 7x + 3 = 0: r+s=72r + s = \frac{7}{2} and rs=32rs = \frac{3}{2}. Therefore, 1r+1s=s+rrs=7/23/2=73\frac{1}{r} + \frac{1}{s} = \frac{s + r}{rs} = \frac{7/2}{3/2} = \frac{7}{3}. Choice A results from incorrectly using rsr+s=3/27/2=37\frac{rs}{r+s} = \frac{3/2}{7/2} = \frac{3}{7}. Choice C comes from the error 7/23=76\frac{7/2}{3} = \frac{7}{6}, using cc instead of ca\frac{c}{a} for the product. Choice D results from using 7/27/3=32\frac{7/2}{7/3} = \frac{3}{2}, confusing the relationships between coefficients and roots.

Question 11

A quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c passes through the points (1,0)(1, 0), (3,0)(3, 0), and (0,k)(0, k) where k>0k > 0. If the discriminant of f(x)=0f(x) = 0 is 36, what is the value of kk?

  1. k=3k = 3
  2. k=6k = 6
  3. k=9k = 9 (correct answer)
  4. k=12k = 12
Explanation: Since f(x)f(x) passes through (1,0)(1, 0) and (3,0)(3, 0), these are the roots of the quadratic. Therefore, f(x)=a(x−1)(x−3)=a(x2−4x+3)=ax2−4ax+3af(x) = a(x-1)(x-3) = a(x^2 - 4x + 3) = ax^2 - 4ax + 3a. So a=aa = a, b=−4ab = -4a, and c=3ac = 3a. Since f(0)=kf(0) = k, we have c=3a=kc = 3a = k, so a=k3a = \frac{k}{3}. The discriminant is Δ=b2−4ac=(−4a)2−4a(3a)=16a2−12a2=4a2\Delta = b^2 - 4ac = (-4a)^2 - 4a(3a) = 16a^2 - 12a^2 = 4a^2. Given that Δ=36\Delta = 36, we have 4a2=364a^2 = 36, so a2=9a^2 = 9, giving a=±3a = \pm 3. Since k=3ak = 3a and k>0k > 0, we need a>0a > 0, so a=3a = 3. Therefore, k=3a=3(3)=9k = 3a = 3(3) = 9. Let's verify: f(x)=3(x−1)(x−3)=3(x2−4x+3)=3x2−12x+9f(x) = 3(x-1)(x-3) = 3(x^2 - 4x + 3) = 3x^2 - 12x + 9. The discriminant is (−12)2−4(3)(9)=144−108=36(-12)^2 - 4(3)(9) = 144 - 108 = 36 ✓. Also, f(0)=9=kf(0) = 9 = k ✓, and f(1)=f(3)=0f(1) = f(3) = 0 ✓. Option A (k=3k = 3) would give a=1a = 1 and discriminant 4(1)2=4≠364(1)^2 = 4 \neq 36. Option B (k=6k = 6) would give a=2a = 2 and discriminant 4(2)2=16≠364(2)^2 = 16 \neq 36. Option D (k=12k = 12) would give a=4a = 4 and discriminant 4(4)2=64≠364(4)^2 = 64 \neq 36.