Answer:
Solution:
1. [tex]x^{2}[/tex] + 4x + 3 = 0, then: a = 1, b= 4, c= 3 using the discriminant formula, we have
[tex]b^{2}[/tex]- 4ac = [tex]4^{2}[/tex] - 4(1)(3)
= 16 - 12
= 4, since the the discriminant is greater than 0 which is 4 and a perfect square, the nature of the roots are: rational number, not equal.
Checked:
[tex]x^{2}[/tex] + 4x + 3 = 0
(x + 1)(x+3) = 0
x +1 =0, x + 3 = 0
x = -1, x = -3, rational number, not equal.
2. [tex]x^{2}[/tex] -5x + 4 =0
Solution:
a= 1, b = -5, c = 4
[tex]b^{2}[/tex] - 4ac = [tex](-5)^{2}[/tex] - 4(1)(4)
= 25 - 16
= 9, since the the discriminant is greater than 0 which is 9 and a perfect square, the nature of the roots are: rational number, not equal.
Checked:
[tex]x^{2} - 5x[/tex] + 4 = 0
(x -1)(x-4) = 0
x - 1 = 0, x - 4 = 0
x = 1, x = 4,rational number, not equal.
3. [tex]x^{2}[/tex] + 7 = 0
Solution:
a = 1, b = 0, c = 7
then,
[tex]b^{2}[/tex] - 4ac = [tex]0^{2}[/tex] - 4(1)(7)
= 0 -28
= -28
Since the discriminant is less than 0 which is -28, the quadratic equation
[tex]x^{2}[/tex] + 7 = 0 has no real roots.
Checked:
[tex]x^{2} + 7 = 0[/tex]
[tex]x^{2}[/tex] = -7
x = [tex]\sqrt{-7}[/tex]
x = [tex]\sqrt{-7}[/tex], no real roots
4. [tex]4x^{2} -4x + 1 = 0[/tex]
Solution:
a = 4, b = -4 c = 1
[tex]b^{2} -4ac[/tex] = [tex](-4)^{2}[/tex] - 4(4)(1)
= 16 - 16
= 0
Since the discriminant is 0, therefore the nature are;
real numbers, equal.
Checked:
[tex]4x^{2} -4x + 1 = 0[/tex]
(2x - 1)(2x - 1) = 0
2x - 1 = 0, 2x - 1 = 0
2x = 1, 2x = 1
[tex]\frac{2x}{2}[/tex] = [tex]\frac{1}{2}[/tex] , [tex]\frac{2x}{2}[/tex] = [tex]\frac{1}{2}[/tex]
real numbers, equal.
Sorry, I cannot answer number 5 because some numbers are missing, at least sa ngayon you have already answers that may help you or others to address their problem.
Salamat!
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Step-by-step explanation:
