Quadratika!
Find the positive value of p such that the quadratic equation p x 2 − 1 2 x + 4 = 0 has only one solution.
The answer is 9.
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13 solutions
Good work, upvoted!
nice solution...
I used trial and error and got p=5 and x=2; (5)(2)^2-12(2)+4=0 But, whatever lol
I thought 0, because there was NOTHING saying that it must be a quadratic equation at last. Then, if P is equal to 0 we have a simple equation -12x+4=0
Another thing is "the quadratic equation ALWAYS has 2 solutions" but, sometimes, they're both equal.
I believe the aswer to the question MUST not be 9, because, if it was, then x' is 4/9 and x'' is 4/9.
If P is 0, then there is only one X= 1/3
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Sorry for late reply. But I see the question has been edited. Earlier it was mentioned that p is not 0.
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It says 'positive value'. Zero is not positive anymore... LOL
Method without using the discriminant: To force a unique solution , the L.H.S must be a perfect square. Recall ( a − b ) 2 = a 2 − 2 a b + b 2 . So observe the middle term. Note that − 1 2 x = − 2 ( 3 x ) ( 2 ) .Since we already have 2 2 = 4 for the L.H.S to be a perfect square , clearly p x 2 = ( 3 x ) 2 = 9 x 2 ⇒ p = 9 .
I thought 0, because there was NOTHING saying that it must be a quadratic equation at last. Then, if P is equal to 0 we have a simple equation -12x+4=0
Another thing is "the quadratic equation ALWAYS has 2 solutions" but, sometimes, they're both equal.
I believe the aswer to the question MUST not be 9, because, if it was, then x' is 4/9 and x'' is 4/9.
If P is 0, then there is only one X= 1/3
Log in to reply
Belief has no place in basic mathematics.
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Every time we have a quadratic equation, we have two roots. In the example, both are equal. TWO both duas dos A Priceless world is better.
For one solution of a quadtric equation we have discriminant equal to 0 . therefore 12^2-16p = 0 ; =》p=9
x=-b/2a gives the x value for the vertex, once you find x, plug that in, chug to find P as the equation is set equal to zero, solve for P
1 2 2 − 4 ( p ) ( 4 ) = 0
4 2 p = 1 2 2
p = ( 4 1 2 ) 2 = 3 2 = 9
For a quadratic equation of the form y= ax^2 + bx+c, for there only to be one solution (when y=0), the discriminant b^2 -4ac must equal zero.
With the knowledge of b,a,c (a being p). you then have a linear equation of 144-16p=0, then rearranging and solving for p you find that p=9.
I thought 0, because there was NOTHING saying that it must be a quadratic equation at last. Then, if P is equal to 0 we have a simple equation -12x+4=0
Another thing is "the quadratic equation ALWAYS has 2 solutions" but, sometimes, they're both equal.
I believe the aswer to the question MUST not be 9, because, if it was, then x' is 4/9 and x'' is 4/9.
If P is 0, then there is only one X= 1/3
For one solution, the discriminant would have to be zero, so \b^2 = 4ac, or (-12)^2 = 4p(4), or 144 = 16p, so p = 9.
Since the equation has only one solution, we can write the equation; p x 2 − 1 2 x + 4 = ( a x + b ) 2 p x 2 − 1 2 x + 4 = a 2 x 2 + 2 a b x + b 2 .
From identity above, we can have b = ± 2 , a = ∓ 3 and p = 9 .
Take the first derivative of eqn. 1 then find px, 2px - 12 = 0 px = 6 From eqn. 1, find x px^2 - 12x + 4 = 0 px(x) - 12x + 4 = 0 6x - 12x + 4 = 0 x = 2/3 then find p px = 6 p(2/3) = 6 p = 9
(3x - 2)(3x -2)
One soln.
P=9
Please explain more
I agree with @Abdur Rehman Zahid . You just proved by substitution that there is only one solution. Is it necessary for other quadratic polynomials There will exist factors (3x-2)(3x-2)?
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See my solution. I think it completes the expected portion :)
THE ANSWER SHOULD BE ZERO SINCE THEY HAVE THE POLYNOMIAL HAS ONLY ONE SOLUTION WHICH IS ONLY POSSIBLE FOR LINEAR EQUATIONS AND NOT FOR QUADRATIC SINCE EVEN THOUGHT THE QUADRATIC POLYNOMIAL HAS ONE VALUE THEY ARE CONSIDERED AS TWO DIFFERENT EQUATIONS AND NOT SINGLE
For the given equation to have one solution, i.e, two identical roots, it's discriminant must be equal to zero. Writing the discriminant and performing some elementary manipulations would yield the value of " P " which is 6.
Discriminant =0 b^2=4ac so p=9
I thought 0, because there was NOTHING saying that it must be a quadratic equation at last. Then, if P is equal to 0 we have a simple equation -12x+4=0
Another thing is "the quadratic equation ALWAYS has 2 solutions" but, sometimes, they're both equal.
I believe the aswer to the question MUST not be 9, because, if it was, then x' is 4/9 and x'' is 4/9.
If P is 0, then there is only one X= 1/3
For a quadratic equation of the form a x 2 + b x + c = 0 to have only 1 solution, then its discriminant b 2 − 4 a c must equal 0. Substituting the values from the given quadratic into b 2 − 4 a c = 0 gives us ( − 1 2 ) 2 − 4 ( p ) ( 4 ) = 0
1 4 4 − 1 6 p = 0 ⟹ p = 9