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Asked by Jogendra Last Modified
As a seasoned tutor registered on UrbanPro, I understand the importance of clarity and precision in solving mathematical problems. Let's tackle this question step by step.
Given the quadratic equation x2+px+8=0x2+px+8=0, we need to compute the value of pp such that the difference of the roots of the equation is 2.
The difference of the roots of a quadratic equation ax2+bx+c=0ax2+bx+c=0 is given by the formula:
Δ=b2−4acΔ=b2−4ac
For our equation x2+px+8=0x2+px+8=0, we have a=1a=1, b=pb=p, and c=8c=8.
Now, we can plug these values into the formula for the difference of the roots:
Δ=p2−4(1)(8)Δ=p2−4(1)(8)
Δ=p2−32Δ=p2−32
Given that the difference of the roots is 2, we can set ΔΔ equal to 2:
2=p2−322=p2−32
To solve for pp, we'll square both sides to eliminate the square root:
4=p2−324=p2−32
p2=4+32p2=4+32
p2=36p2=36
p=±6p=±6
Thus, the value of pp such that the difference of the roots of the equation x2+px+8=0x2+px+8=0 is 2, is p=±6p=±6.
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