if -5 is a root of the quadratic equation 2x^2+px-15=0 and roots of p(x² +x)+k=0 are equal roots,then find the value of p and k
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2x² + px - 15 = 0
if -5 is root of the equation
⇒ When x = -5, the equation will return 0
Find p:
2(-5)² + p (-5) - 15 = 0
50 - 5p - 15 = 0
5p = 35
p = 7
p(x² + x) + k = 0
Since = 7:
7(x² + x) + k = 0
7x² + 7x + k = 0
The equation has equal roots
⇒ the discriminant = 0
Find k:
b² - 4ac = 0
7² - 4(7)(k) = 0
49 - 28k = 0
28k = 49
k = 49 ÷ 29
k = 1.75
Answer: p = 7 and k = 1.75
sourabhsenger101:
thank you so much
You are welcome :)
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Given -5 as the root of eq.
finding the solution
2(-5)²+p(-5)-15=0
50-5p-15=0
5p=35
p=7
p(x²+x)+k=0
As,7
7(x²+x)+k=0
=7x²+7x+k=0
Then k,
b²-4ac=0
7²-4(7)(k)=0
49-28k=0
28k=49
k=1.75
finding the solution
2(-5)²+p(-5)-15=0
50-5p-15=0
5p=35
p=7
p(x²+x)+k=0
As,7
7(x²+x)+k=0
=7x²+7x+k=0
Then k,
b²-4ac=0
7²-4(7)(k)=0
49-28k=0
28k=49
k=1.75
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