Find the range of values of p for which the quadratic equation : px^2 + (p+3) + p = 0 : has real roots
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with the discriminant rules of quadratics for and quadratic to have real roots,
b^2 - 4(a)(c) must be >= 0
so (p+3)^2 -4(p)(p) >=0
-3p^2 +6p +9 >= 0
p^2 -2p -3 <= 0
(p-3)(p+1) <=0
giving endpoints of -1 and 3
-1
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b^2 - 4(a)(c) must be >= 0
so (p+3)^2 -4(p)(p) >=0
-3p^2 +6p +9 >= 0
p^2 -2p -3 <= 0
(p-3)(p+1) <=0
giving endpoints of -1 and 3
-1
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Well that's the same as
0= px^2 + 0x +(2p + 3)
so x has real roots when b^2 - 4ac >= 0
b in this case is 0, a is p and c is 2p+3
so -8p^2 - 12p > = 0
2p^2 + 3p >= 0
2 (p^2 + 3p/2 + 9/16) -9/8 >= 0
2 (p + 3/4)^2 - 9/8 >=0
so that's a parabola open up and the vertex is below the x-axis, we set the parabola equal to zero, so the points where the parabola crosses zero are: 0 and -6/8
so p>= 0 or p<= -6/8
0= px^2 + 0x +(2p + 3)
so x has real roots when b^2 - 4ac >= 0
b in this case is 0, a is p and c is 2p+3
so -8p^2 - 12p > = 0
2p^2 + 3p >= 0
2 (p^2 + 3p/2 + 9/16) -9/8 >= 0
2 (p + 3/4)^2 - 9/8 >=0
so that's a parabola open up and the vertex is below the x-axis, we set the parabola equal to zero, so the points where the parabola crosses zero are: 0 and -6/8
so p>= 0 or p<= -6/8
1
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