Question

Use the discriminant to determine what type of roots the equations will have, and categorize the equations according to their roots.
two distinct roots, One repeated root, two complex roots



x^2 − 4x + 2 = 0

5x^2 − 2x + 3 = 0

2x^2 + x − 6 = 0

13x^2 − 4 = 0

x^2 − 6x + 9 = 0

x^2 − 8x + 16 = 0

4x^2 + 11 = 0

Answer 1

Answer:

For the equation x2 − 4x + 2 = 0, the discriminant is (-4)2 − 4(1)(2) = 8. Since the discriminant is positive, it has two distinct real roots.

For the equation 5x2 − 2x + 3 = 0, the discriminant is (-2)2 − 4(5)(3) = -56. Since the discriminant is negative, it has two complex roots.

For the equation 2x2 + x − 6 = 0, the discriminant is (1)2 − 4(2)(-6) = 49. Since the discriminant is positive, it has two distinct real roots.

For the equation 13x2 − 4 = 0, the discriminant is (0)2 − 4(13)(-4) = 208. Since the discriminant is positive, it has two distinct real roots.

For the equation x2 − 6x + 9 = 0, the discriminant is (-6)2 − 4(1)(9) = 0. Since the discriminant is zero, it has one repeated root.

For the equation x2 − 8x + 16 = 0, the discriminant is (-8)2 − 4(1)(16) = 0. Since the discriminant is zero, it has one repeated root.

For the equation 4x2 + 11 = 0, the discriminant is (0)2 − 4(4)(11) = -176. Since the discriminant is negative, it has two complex roots.

Answer 2

Step-by-step explanation:

The discriminant of the quadratic equation $ax^2+bx+c=0$:

$\Delta=b^2-4ac$

If Δ < 0, then the equation has two complex roots $x=\dfrac{-b\pm\sqrt\Delta}{2a}$

If Δ = 0, then the equation has one repeated root $x=\dfrac{-b}{2a}[/tex If Δ > 0, then the equation has two discint roots [tex]x=\dfrac{-b\pm\sqrt\Delta}{2a}$

$1.\ x^2-4x+2=0\\\\a=1,\ b=-4,\ c=2\\\\\Delta=(-4)^2-4(1)(2)=16-8=8>0,\ \bold{two\ distinct\ roots}\\\sqrt\Delta=\sqrt8=\sqrt{4\cdot2}=2\sqrt2\\\\x=\dfrac{-(-4)\pm2\sqrt2}{2(1)}=\dfrac{4\pm2\sqrt2}{2}=2\pm\sqrt2\\\\==============================\\\\2.\ 5x^2-2x+3=0\\\\a=5,\ b=-2,\ c=3\\\\\Delta=(-2)^2-4(5)(3)=4-60=-56$

$3.\ 2x^2+x-6=0\\\\a=2,\ b=1,\ c=-6\\\\\Delta=1^2-4(2)(-6)=1+48=49>0,\ \bold{two\ distinct\ roots}\\\sqrt\Delta=\sqrt{49}=7\\\\x=\dfrac{-1\pm7}{(2)(2)}\\\\x_1=\dfrac{-8}{4}=-2,\ x_2=\dfrac{6}{4}=\dfrac{3}{2}\\\\==============================\\\\4.\ 13x^2-4=0\qquad\text{add 4 to both sides}\\\\13x^2=4\qquad\text{divide both sides by 13}\\\\x^2=\dfrac{4}{13}\to x=\pm\sqrt{\dfrac{4}{13}},\ \bold{two\ distinct\ roots}\\\\==============================$

$5.\ x^2-6x+16=0\\\\a=1,\ b=-6,\ c=16\\\\\Delta=(-6)^2-4(1)(16)=36-64=-28$

$7.\ 4x^2+11=0\qquad\text{subtract 11 from both sides}\\\\4x^2=-11\qquad\text{divide both sides by 4}\\\\x^2=-\dfrac{11}{4}\to x=\pm\sqrt{-\dfrac{11}{4}}\\\\x=\pm\dfrac{\sqrt{11}}{2}\ i,\ \bold{two\ complex\ roots}$