Practice
Quadratic Formula and the Discriminant Practice
Fifty original questions on coefficients, exact roots, discriminants, root count, parameters, graph behavior, and root relationships.
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Question 1
Explanation
Standard form is \(ax^2+bx+c=0\). Reading each signed coefficient gives \(a=3\), \(b=-7\), and \(c=2\).
- Method
Write the sign with each coefficient before using the quadratic formula.
- Conclusion
Standard form is \(ax^2+bx+c=0\). Reading each signed coefficient gives \(a=3\), \(b=-7\), and \(c=2\).
Question 2
Explanation
Standard form is \(ax^2+bx+c=0\). Reading each signed coefficient gives \(a=-2\), \(b=5\), and \(c=-9\).
- Method
Write the sign with each coefficient before using the quadratic formula.
- Conclusion
Standard form is \(ax^2+bx+c=0\). Reading each signed coefficient gives \(a=-2\), \(b=5\), and \(c=-9\).
Question 3
Explanation
Standard form is \(ax^2+bx+c=0\). Reading each signed coefficient gives \(a=4\), \(b=0\), and \(c=-11\).
- Method
Write the sign with each coefficient before using the quadratic formula.
- Conclusion
Standard form is \(ax^2+bx+c=0\). Reading each signed coefficient gives \(a=4\), \(b=0\), and \(c=-11\).
Question 4
Explanation
Standard form is \(ax^2+bx+c=0\). Reading each signed coefficient gives \(a=1\), \(b=-12\), and \(c=8\).
- Method
Write the sign with each coefficient before using the quadratic formula.
- Conclusion
Standard form is \(ax^2+bx+c=0\). Reading each signed coefficient gives \(a=1\), \(b=-12\), and \(c=8\).
Question 5
Explanation
Standard form is \(ax^2+bx+c=0\). Reading each signed coefficient gives \(a=-5\), \(b=-3\), and \(c=6\).
- Method
Write the sign with each coefficient before using the quadratic formula.
- Conclusion
Standard form is \(ax^2+bx+c=0\). Reading each signed coefficient gives \(a=-5\), \(b=-3\), and \(c=6\).
Question 6
Explanation
Use \(D=b^2-4ac\): \(D=(5)^2-4(1)(2)=17\).
- Method
Calculate the squared linear coefficient and signed product separately.
- Conclusion
Use \(D=b^2-4ac\): \(D=(5)^2-4(1)(2)=17\).
Question 7
Explanation
Use \(D=b^2-4ac\): \(D=(-6)^2-4(2)(1)=28\).
- Method
Calculate the squared linear coefficient and signed product separately.
- Conclusion
Use \(D=b^2-4ac\): \(D=(-6)^2-4(2)(1)=28\).
Question 8
Explanation
Use \(D=b^2-4ac\): \(D=(4)^2-4(3)(-5)=76\).
- Method
Calculate the squared linear coefficient and signed product separately.
- Conclusion
Use \(D=b^2-4ac\): \(D=(4)^2-4(3)(-5)=76\).
Question 9
Explanation
Use \(D=b^2-4ac\): \(D=(8)^2-4(-1)(3)=76\).
- Method
Calculate the squared linear coefficient and signed product separately.
- Conclusion
Use \(D=b^2-4ac\): \(D=(8)^2-4(-1)(3)=76\).
Question 10
Explanation
Use \(D=b^2-4ac\): \(D=(-2)^2-4(4)(7)=-108\).
- Method
Calculate the squared linear coefficient and signed product separately.
- Conclusion
Use \(D=b^2-4ac\): \(D=(-2)^2-4(4)(7)=-108\).
Question 11
Explanation
The discriminant is \(36\), which is positive. Therefore there are 2 real roots, matching the number of horizontal-axis intersections.
- Method
Connect the discriminant sign to the number of horizontal-axis intersections.
- Conclusion
The discriminant is \(36\), which is positive. Therefore there are 2 real roots, matching the number of horizontal-axis intersections.
Question 12
Explanation
The discriminant is \(0\), which is zero. Therefore there is 1 real root, matching the number of horizontal-axis intersections.
- Method
Connect the discriminant sign to the number of horizontal-axis intersections.
- Conclusion
The discriminant is \(0\), which is zero. Therefore there is 1 real root, matching the number of horizontal-axis intersections.
Question 13
Explanation
The discriminant is \(-36\), which is negative. Therefore there are 0 real roots, matching the number of horizontal-axis intersections.
- Method
Connect the discriminant sign to the number of horizontal-axis intersections.
- Conclusion
The discriminant is \(-36\), which is negative. Therefore there are 0 real roots, matching the number of horizontal-axis intersections.
Question 14
Explanation
The discriminant is \(36\), which is positive. Therefore there are 2 real roots, matching the number of horizontal-axis intersections.
- Method
Connect the discriminant sign to the number of horizontal-axis intersections.
- Conclusion
The discriminant is \(36\), which is positive. Therefore there are 2 real roots, matching the number of horizontal-axis intersections.
Question 15
Explanation
The discriminant is \(0\), which is zero. Therefore there is 1 real root, matching the number of horizontal-axis intersections.
- Method
Connect the discriminant sign to the number of horizontal-axis intersections.
- Conclusion
The discriminant is \(0\), which is zero. Therefore there is 1 real root, matching the number of horizontal-axis intersections.
Question 16
Explanation
Substituting the signed coefficients into \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) produces \(x=-4\) and \(x=7\). Both values make the original quadratic zero.
- Method
Compute the discriminant first, then simplify the plus and minus branches separately.
- Conclusion
Substituting the signed coefficients into \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) produces \(x=-4\) and \(x=7\). Both values make the original quadratic zero.
Question 17
Explanation
Substituting the signed coefficients into \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) produces \(x=-3\) and \(x=5\). Both values make the original quadratic zero.
- Method
Compute the discriminant first, then simplify the plus and minus branches separately.
- Conclusion
Substituting the signed coefficients into \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) produces \(x=-3\) and \(x=5\). Both values make the original quadratic zero.
Question 18
Explanation
Substituting the signed coefficients into \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) produces \(x=2\) and \(x=8\). Both values make the original quadratic zero.
- Method
Compute the discriminant first, then simplify the plus and minus branches separately.
- Conclusion
Substituting the signed coefficients into \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) produces \(x=2\) and \(x=8\). Both values make the original quadratic zero.
Question 19
Explanation
Substituting the signed coefficients into \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) produces \(x=-2\) and \(x=4\). Both values make the original quadratic zero.
- Method
Compute the discriminant first, then simplify the plus and minus branches separately.
- Conclusion
Substituting the signed coefficients into \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) produces \(x=-2\) and \(x=4\). Both values make the original quadratic zero.
Question 20
Explanation
Substituting the signed coefficients into \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) produces \(x=1\) and \(x=6\). Both values make the original quadratic zero.
- Method
Compute the discriminant first, then simplify the plus and minus branches separately.
- Conclusion
Substituting the signed coefficients into \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) produces \(x=1\) and \(x=6\). Both values make the original quadratic zero.
Question 21
Explanation
Using \(a=1\), \(b=2\), and \(c=-5\), the discriminant is \(24\). Substitution and radical simplification give \(x=-1\pm\sqrt6\). Both branches satisfy the original equation.
- Method
Keep the entire numerator over the denominator and reduce common factors after simplifying the radical.
- Conclusion
Using \(a=1\), \(b=2\), and \(c=-5\), the discriminant is \(24\). Substitution and radical simplification give \(x=-1\pm\sqrt6\). Both branches satisfy the original equation.
Question 22
Explanation
Using \(a=2\), \(b=4\), and \(c=-3\), the discriminant is \(40\). Substitution and radical simplification give \(x=-1\pm\frac{\sqrt{10}}2\). Both branches satisfy the original equation.
- Method
Keep the entire numerator over the denominator and reduce common factors after simplifying the radical.
- Conclusion
Using \(a=2\), \(b=4\), and \(c=-3\), the discriminant is \(40\). Substitution and radical simplification give \(x=-1\pm\frac{\sqrt{10}}2\). Both branches satisfy the original equation.
Question 23
Explanation
Using \(a=3\), \(b=-6\), and \(c=-2\), the discriminant is \(60\). Substitution and radical simplification give \(x=1\pm\frac{\sqrt{15}}3\). Both branches satisfy the original equation.
- Method
Keep the entire numerator over the denominator and reduce common factors after simplifying the radical.
- Conclusion
Using \(a=3\), \(b=-6\), and \(c=-2\), the discriminant is \(60\). Substitution and radical simplification give \(x=1\pm\frac{\sqrt{15}}3\). Both branches satisfy the original equation.
Question 24
Explanation
Using \(a=1\), \(b=-4\), and \(c=-1\), the discriminant is \(20\). Substitution and radical simplification give \(x=2\pm\sqrt5\). Both branches satisfy the original equation.
- Method
Keep the entire numerator over the denominator and reduce common factors after simplifying the radical.
- Conclusion
Using \(a=1\), \(b=-4\), and \(c=-1\), the discriminant is \(20\). Substitution and radical simplification give \(x=2\pm\sqrt5\). Both branches satisfy the original equation.
Question 25
Explanation
Using \(a=2\), \(b=-2\), and \(c=-5\), the discriminant is \(44\). Substitution and radical simplification give \(x=\frac{1\pm\sqrt{11}}2\). Both branches satisfy the original equation.
- Method
Keep the entire numerator over the denominator and reduce common factors after simplifying the radical.
- Conclusion
Using \(a=2\), \(b=-2\), and \(c=-5\), the discriminant is \(44\). Substitution and radical simplification give \(x=\frac{1\pm\sqrt{11}}2\). Both branches satisfy the original equation.
Question 26
Explanation
Exactly one real solution requires \(D=0\): \(k^2-4(4)=0\), so \(k^2=16\) and \(k=\pm4\).
- Method
Tangency corresponds to a zero discriminant; solve the resulting parameter equation completely.
- Conclusion
Exactly one real solution requires \(D=0\): \(k^2-4(4)=0\), so \(k^2=16\) and \(k=\pm4\).
Question 27
Explanation
Exactly one real solution requires \(D=0\): \(k^2-4(9)=0\), so \(k^2=36\) and \(k=\pm6\).
- Method
Tangency corresponds to a zero discriminant; solve the resulting parameter equation completely.
- Conclusion
Exactly one real solution requires \(D=0\): \(k^2-4(9)=0\), so \(k^2=36\) and \(k=\pm6\).
Question 28
Explanation
Exactly one real solution requires \(D=0\): \(k^2-4(16)=0\), so \(k^2=64\) and \(k=\pm8\).
- Method
Tangency corresponds to a zero discriminant; solve the resulting parameter equation completely.
- Conclusion
Exactly one real solution requires \(D=0\): \(k^2-4(16)=0\), so \(k^2=64\) and \(k=\pm8\).
Question 29
Explanation
Exactly one real solution requires \(D=0\): \(k^2-4(25)=0\), so \(k^2=100\) and \(k=\pm10\).
- Method
Tangency corresponds to a zero discriminant; solve the resulting parameter equation completely.
- Conclusion
Exactly one real solution requires \(D=0\): \(k^2-4(25)=0\), so \(k^2=100\) and \(k=\pm10\).
Question 30
Explanation
Exactly one real solution requires \(D=0\): \(k^2-4(36)=0\), so \(k^2=144\) and \(k=\pm12\).
- Method
Tangency corresponds to a zero discriminant; solve the resulting parameter equation completely.
- Conclusion
Exactly one real solution requires \(D=0\): \(k^2-4(36)=0\), so \(k^2=144\) and \(k=\pm12\).
Question 31
Explanation
Direct calculation gives \(D=(-4)^2-4(1)(3)=4\). Its sign predicts that the graph crosses the horizontal axis twice, exactly as rendered.
- Method
Use coefficients to calculate the discriminant, then use the graph as a consistency check.
- Conclusion
Direct calculation gives \(D=(-4)^2-4(1)(3)=4\). Its sign predicts that the graph crosses the horizontal axis twice, exactly as rendered.
Question 32
Explanation
Direct calculation gives \(D=(2)^2-4(1)(1)=0\). Its sign predicts that the graph touches the horizontal axis once, exactly as rendered.
- Method
Use coefficients to calculate the discriminant, then use the graph as a consistency check.
- Conclusion
Direct calculation gives \(D=(2)^2-4(1)(1)=0\). Its sign predicts that the graph touches the horizontal axis once, exactly as rendered.
Question 33
Explanation
Direct calculation gives \(D=(0)^2-4(2)(6)=-48\). Its sign predicts that the graph does not meet the horizontal axis, exactly as rendered.
- Method
Use coefficients to calculate the discriminant, then use the graph as a consistency check.
- Conclusion
Direct calculation gives \(D=(0)^2-4(2)(6)=-48\). Its sign predicts that the graph does not meet the horizontal axis, exactly as rendered.
Question 34
Explanation
Direct calculation gives \(D=(-2)^2-4(-1)(8)=36\). Its sign predicts that the graph crosses the horizontal axis twice, exactly as rendered.
- Method
Use coefficients to calculate the discriminant, then use the graph as a consistency check.
- Conclusion
Direct calculation gives \(D=(-2)^2-4(-1)(8)=36\). Its sign predicts that the graph crosses the horizontal axis twice, exactly as rendered.
Question 35
Explanation
Direct calculation gives \(D=(6)^2-4(3)(5)=-24\). Its sign predicts that the graph does not meet the horizontal axis, exactly as rendered.
- Method
Use coefficients to calculate the discriminant, then use the graph as a consistency check.
- Conclusion
Direct calculation gives \(D=(6)^2-4(3)(5)=-24\). Its sign predicts that the graph does not meet the horizontal axis, exactly as rendered.
Question 36
Explanation
For \(ax^2+bx+c=0\), the sum is \(-\frac ba=-\frac{-10}{2}=5\).
- Method
Use the root-sum relationship when individual roots are not requested.
- Conclusion
For \(ax^2+bx+c=0\), the sum is \(-\frac ba=-\frac{-10}{2}=5\).
Question 37
Explanation
For \(ax^2+bx+c=0\), the sum is \(-\frac ba=-\frac{12}{3}=-4\).
- Method
Use the root-sum relationship when individual roots are not requested.
- Conclusion
For \(ax^2+bx+c=0\), the sum is \(-\frac ba=-\frac{12}{3}=-4\).
Question 38
Explanation
For \(ax^2+bx+c=0\), the sum is \(-\frac ba=-\frac{14}{-2}=7\).
- Method
Use the root-sum relationship when individual roots are not requested.
- Conclusion
For \(ax^2+bx+c=0\), the sum is \(-\frac ba=-\frac{14}{-2}=7\).
Question 39
Explanation
For \(ax^2+bx+c=0\), the sum is \(-\frac ba=-\frac{-15}{5}=3\).
- Method
Use the root-sum relationship when individual roots are not requested.
- Conclusion
For \(ax^2+bx+c=0\), the sum is \(-\frac ba=-\frac{-15}{5}=3\).
Question 40
Explanation
For \(ax^2+bx+c=0\), the sum is \(-\frac ba=-\frac{-20}{-4}=-5\).
- Method
Use the root-sum relationship when individual roots are not requested.
- Conclusion
For \(ax^2+bx+c=0\), the sum is \(-\frac ba=-\frac{-20}{-4}=-5\).
Question 41
Explanation
The product of roots is \(\frac ca=\frac{-12}{2}=-6\). Solving for each root is unnecessary.
- Method
Match the requested relationship to the coefficient formula before using the full quadratic formula.
- Conclusion
The product of roots is \(\frac ca=\frac{-12}{2}=-6\). Solving for each root is unnecessary.
Question 42
Explanation
The product of roots is \(\frac ca=\frac{20}{4}=5\). Solving for each root is unnecessary.
- Method
Match the requested relationship to the coefficient formula before using the full quadratic formula.
- Conclusion
The product of roots is \(\frac ca=\frac{20}{4}=5\). Solving for each root is unnecessary.
Question 43
Explanation
The product of roots is \(\frac ca=\frac{15}{-3}=-5\). Solving for each root is unnecessary.
- Method
Match the requested relationship to the coefficient formula before using the full quadratic formula.
- Conclusion
The product of roots is \(\frac ca=\frac{15}{-3}=-5\). Solving for each root is unnecessary.
Question 44
Explanation
The product of roots is \(\frac ca=\frac{-25}{5}=-5\). Solving for each root is unnecessary.
- Method
Match the requested relationship to the coefficient formula before using the full quadratic formula.
- Conclusion
The product of roots is \(\frac ca=\frac{-25}{5}=-5\). Solving for each root is unnecessary.
Question 45
Explanation
The product of roots is \(\frac ca=\frac{-14}{-2}=7\). Solving for each root is unnecessary.
- Method
Match the requested relationship to the coefficient formula before using the full quadratic formula.
- Conclusion
The product of roots is \(\frac ca=\frac{-14}{-2}=7\). Solving for each root is unnecessary.
Question 46
Explanation
The linear coefficient is negative, so b must be recorded with a negative sign. Standard form, a separately computed discriminant, and substitution of final roots provide independent checks.
- Method
Write signed coefficients and the discriminant on separate lines before substitution.
- Conclusion
The linear coefficient is negative, so b must be recorded with a negative sign. Standard form, a separately computed discriminant, and substitution of final roots provide independent checks.
Question 47
Explanation
The discriminant uses subtraction: b squared minus four times a times c. Standard form, a separately computed discriminant, and substitution of final roots provide independent checks.
- Method
Write signed coefficients and the discriminant on separate lines before substitution.
- Conclusion
The discriminant uses subtraction: b squared minus four times a times c. Standard form, a separately computed discriminant, and substitution of final roots provide independent checks.
Question 48
Explanation
The entire numerator, including the opposite of b, is divided by 2a. Standard form, a separately computed discriminant, and substitution of final roots provide independent checks.
- Method
Write signed coefficients and the discriminant on separate lines before substitution.
- Conclusion
The entire numerator, including the opposite of b, is divided by 2a. Standard form, a separately computed discriminant, and substitution of final roots provide independent checks.
Question 49
Explanation
A zero discriminant gives one repeated real root. Standard form, a separately computed discriminant, and substitution of final roots provide independent checks.
- Method
Write signed coefficients and the discriminant on separate lines before substitution.
- Conclusion
A zero discriminant gives one repeated real root. Standard form, a separately computed discriminant, and substitution of final roots provide independent checks.
Question 50
Explanation
Use negative b over a to answer directly and reduce arithmetic risk. Standard form, a separately computed discriminant, and substitution of final roots provide independent checks.
- Method
Write signed coefficients and the discriminant on separate lines before substitution.
- Conclusion
Use negative b over a to answer directly and reduce arithmetic risk. Standard form, a separately computed discriminant, and substitution of final roots provide independent checks.
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