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Math

Math Snap

PROBLEM

Find the range of values of pp such that both 3x2+2px+3=03x^{2}+2px+3=0 and x2+4xp=0x^{2}+4x-p=0 have real roots.

STEP 1

Assumptions1. The two equations given are quadratic equations.
. A quadratic equation ax+bx+c=0ax^ + bx + c =0 has real roots if the discriminant b4acb^ -4ac is greater than or equal to zero.
3. We need to find the range of values of pp for which both equations have real roots.

STEP 2

Let's start with the first equation. The discriminant of the first equation is 1=(2p)241 = (2p)^2 -4**.

STEP 3

implify the discriminant of the first equation.
1=p2361 =p^2 -36

STEP 4

For the first equation to have real roots, the discriminant 11 must be greater than or equal to zero. So, we have the inequality4p23604p^2 -36 \geq0

STEP 5

olve the inequality for pp.
4p2364p^2 \geq36p29p^2 \geq9p3orp3p \geq3 \quad or \quad p \leq -3

STEP 6

Now, let's move to the second equation. The discriminant of the second equation is 2=(4)24<em>1</em>(p)2 = (4)^2 -4<em>1</em>(-p).

STEP 7

implify the discriminant of the second equation.
2=16+4p2 =16 +4p

STEP 8

For the second equation to have real roots, the discriminant 22 must be greater than or equal to zero. So, we have the inequality16+4p016 +4p \geq0

STEP 9

olve the inequality for pp.
4p164p \geq -16p4p \geq -4

SOLUTION

The range of pp for which both equations have real roots is the intersection of the ranges found in steps5 and9. So, we havep3or4p3p \geq3 \quad or \quad -4 \leq p \leq -3The range of values of pp for which both the equations 3x2+2px+3=03 x^{2}+2 p x+3=0 and x2+4xp=0x^{2}+4 x-p=0 have real roots is p3p \geq3 and 4p3-4 \leq p \leq -3.

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