Answers: Introduction to quadratic equations
These are the answers to Questions: Introduction to quadratic equations.
Please attempt the questions before reading these answers!
Q1
For each of the quadratic equations below, identify the variable and the coefficients \(a,b,c\).
1.1. For \(x^2 - 7x + 6 = 0\), the variable is \(x\), and the coefficients are \(a = 1,b= -7, c=6\).
1.2. For \(y^2 + 14y + 49 = 0\), the variable is \(y\), and the coefficients are \(a = 1,b= 14, c=49\).
1.3. For \(h^2 - h - 56 = 0\), the variable is \(h\), and the coefficients are \(a = 1,b= -1, c= -56\).
1.4. For \(7y^4 - y^2 = 0\), the variable is \(y^2\), and the coefficients are \(a = 7, b=-1,c=0\).
1.5. For \(5n^2 - 14n + 100 = 0\), the variable is \(n\), and the coefficients are \(a = 5, b=-14,c=100\).
1.6. For \(A^2 - 144 = 0\), the variable is \(A\), and the coefficients are \(a = 1, b=0,c=-144\).
1.7. For \(25M^2= 0\), the variable is \(M\), and the coefficients are \(a = 25, b=0,c=0\).
1.8. For \(e^{2x} - 4e^x + 4 = 0\), the variable is \(e^x\), and the coefficients are \(a = 1, b=-4,c=4\).
1.9. For \(-9s^4 + 3s^2 - 1 = 0\), the variable is \(s^2\), and the coefficients are \(a = -9, b=3,c=-1\).
1.10. For \(2e^{6x} + e^{3x} + 1 = 0\), the variable is \(e^{3x}\), and the coefficients are \(a = 2, b=1,c=1\).
1.11. For \(\cos^2(x) + 4\cos(x) - 4 = 0\), the variable is \(\cos(x)\), and the coefficients are \(a = 1, b=4,c=-4\).
1.12. For \(8x^8 - 4x^4 - 1 = 0\), the variable is \(x^4\), and the coefficients are \(a = 8, b=-4,c=-1\).
Q2
2.1. The discriminant of the equation \(x^2 - 7x + 6 = 0\) is \(D = 25\), and therefore the equation has two distinct real roots.
2.2. The discriminant of the equation \(y^2 + 14y + 49 = 0\) is \(D = 0\), and therefore the equation has one distinct real root.
2.3. The discriminant of the equation \(h^2 - h - 56 = 0\) is \(D = 217\), and therefore the equation has two distinct real roots.
2.4. The discriminant of the equation \(7y^4 - y^2 = 0\) is \(D = 1\), and therefore the equation has two distinct real roots.
2.5. The discriminant of the equation \(5n^2 - 14n + 100 = 0\) is \(D = -1804\), and therefore the equation has no real roots (two distinct complex roots).
2.6. The discriminant of the equation \(A^2 - 144 = 0\) is \(D = 576\), and therefore the equation has two distinct real roots.
2.7. The discriminant of the equation \(25M^2 = 0\) is \(D = 0\), and therefore the equation has one distinct real root.
2.8. The discriminant of the equation \(e^{2x} - 4e^x + 4 = 0\) is \(D = 0\), and therefore the equation has one distinct real root \(r\) in \(e^x\). Whether or not it has a real root in \(x\) depends on whether or not \(r\) is positive. If \(r\) is positive, there is exactly one real root \(x = \ln(r)\); if \(r\) is negative, then there are no real roots.
2.9. The discriminant of the equation \(-9s^4 + 3s^2 - 1 = 0\) is \(D = -27\), and therefore the equation has no real roots. This is true even with \(s^2\) as the variable, as if \(s^2\) is complex then \(s\) must also be complex.
2.10. The discriminant of the equation \(2e^{6x} + e^{3x} + 1 = 0\) is \(D = -7\), and therefore the equation has no real roots. This is true even with \(e^{3x}\) as the variable, as if \(e^{3x}\) is complex then \(x\) must also be complex.
2.11. The discriminant of the equation \(\cos^2(x) + 4\cos(x) - 4 = 0\) is \(D = 32\), and therefore the equation has two distinct real roots \(r_1\) and \(r_2\) in \(\cos(x)\). Whether or not it has a real root in \(x\) depends on whether or not either of the roots is between \(-1\) and \(1\). If both \(r_1\) and \(r_2\) are outside this range, then there are no real roots. If one of \(r_1\) or \(r_2\) is between \(-1\) and \(1\), then there are infinitely many solutions.
2.12. The discriminant of the equation \(8x^8 - 4x^4 - 1 = 0\) is \(D = 48\), and therefore the equation has two distinct real roots \(r_1\) and \(r_2\) in \(x^4\). The amount of real roots depend on the signs of \(r_1\) and \(r_2\).
If \(r_1\) and \(r_2\) are both positive, then there are four real roots in \(x\). This is because \(x^2 = \pm\sqrt{r_1}\) or \(x^2 = \pm\sqrt{r_2}\); square rooting the positive terms gives the roots in \(x\) as \(\pm\sqrt{(\sqrt{r_1})} = \pm\sqrt[4]{r_1}\) and \(\pm\sqrt{(\sqrt{r_2})} = \pm\sqrt[4]{r_2}\). Any other roots must be complex, since you are taking square roots of the negative numbers \(-\sqrt{r_1}\) and \(-\sqrt{r_2}\).
If exactly one of \(r_1\) and \(r_2\) is positive (say \(r_i\)), then there are two real roots in \(x\) given by \(\pm\sqrt[4]{r_i}\). All other roots are complex.
If both \(r_1\) and \(r_2\) are negative, then then there are no real roots in \(x\).
Q3
3.1. Rearranging gives \(x^2 + x - 1 = 0\). The discriminant of this is \(D = 5\), and therefore the equation has two distinct real roots.
3.2. Rearranging gives \(y^2 + 10 = 0\). The discriminant of this is \(D = -40\), and therefore the equation has no real roots (two distinct complex roots).
3.3. Rearranging gives \(4m^2 + 4m + 1 = 0\). The discriminant of this is \(D = 0\), and therefore the equation has one distinct real root.
3.4. Rearranging gives \(t^4 + 1 = 0\). The discriminant of this is \(D = -4\), and therefore the equation has no real roots. This is true even with \(t^2\) as the variable, as if \(t^2\) is complex then \(t\) must also be complex.
3.5. Rearranging gives \(5x^2 - 11x - 1 = 0\). The discriminant of this is \(D = 101\), and therefore the equation has two distinct real roots.
3.6. Rearranging gives \(e^{2x} - 2e^x + 1 = 0\). The discriminant of this is \(D = 0\), and therefore the equation has one distinct real root \(r\) in \(e^x\). Whether or not it has a real root in \(x\) depends on whether or not \(r\) is positive. If \(r\) is positive, there is exactly one real root \(x = \ln(r)\); if \(r\) is negative, then there are no real roots.
Version history and licensing
v1.0: initial version created 04/23 by tdhc.
- v1.1: edited 05/24 by tdhc.