Determine the value of p given quadratic equation has real roots 4^x2+8x-p=0
Sakshi15403:
does it have real and equal roots?
has*
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Answered by
13
HOPE IT WILL HELP YOU... :-)
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it is given that roots are real but it isn't given that roots are equal
yeah,...if it's not equal then it'll be p>-4
greater than 4...
D > 0 n ..edit it
D will greater than 0
But there are two conditions na...! maybe equal or unequal...the inquirer should have clarified it...
Answered by
9
Hi ,
Compare 4x² + 8x - p = 0 with
ax² + bx + c = 0 ,
a = 4 ,b = 8 , c = - p
Discreaminant = D = b² - 4ac = 0
1 ) if roots are real
D > 0
b² - 4ac > 0
8² - 4 × 4 × ( -p )>0
64 + 16 p > 0
16 p > - 64
p >- 64/16
p > - 4
2 )If it has equal real roots .
8² - 4 × 4 × ( - p ) = 0
64 + 16p = 0
16 ( 4 + p ) = 0
4 + p = 0
p = - 4
Therefore ,
If p = -4 , the given equation has real
and equal roots.
I hope this helps you.
: )
Compare 4x² + 8x - p = 0 with
ax² + bx + c = 0 ,
a = 4 ,b = 8 , c = - p
Discreaminant = D = b² - 4ac = 0
1 ) if roots are real
D > 0
b² - 4ac > 0
8² - 4 × 4 × ( -p )>0
64 + 16 p > 0
16 p > - 64
p >- 64/16
p > - 4
2 )If it has equal real roots .
8² - 4 × 4 × ( - p ) = 0
64 + 16p = 0
16 ( 4 + p ) = 0
4 + p = 0
p = - 4
Therefore ,
If p = -4 , the given equation has real
and equal roots.
I hope this helps you.
: )
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