Quite A Tough One To Crack

Algebra Level 1

For the quadratic equation, x^2 - 4x + c = 0 where the equation has only one solution,Find the value of c.

2 5 3 4

Varun Srinivas by varun srinivas , 21, India

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9 solutions

Jithin Mathew
Mar 12, 2014

It's so easy....just find the discriminant,

b2-4 a c...

then solve this equation for =0,thus u get the answer..

16-4 a c=0

==>16=4c ==>c=4

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Quadratic equation are in the form of: a x 2 + b x + c = 0 ax^{2}+bx+c=0

In the equation: x 2 − 4 x + c = 0 x^{2}-4x+c=0

a = 1 , b = − 4 , c = c a=1, b=-4, c=c

Discriminant = b 2 − 4 a c = b^{2}-4ac

( − 4 ) 2 − 4 × 1 × c = 0 (-4)^{2}-4 \times 1 \times c = 0 (Since they are equal roots, Discriminant = 0)

16 − 4 c = 0 16-4c=0

− 4 c = 0 − 16 -4c=0-16

c = − 16 − 4 = 4 c=\frac{-16}{-4} = 4

Thus, c = 4 c=\boxed{4} .

Saurabh Mallik - 7 years, 2 months ago
Rashi Agarwal
Mar 10, 2014

x^2-4x+c=0 will have only one solution when it is a square number. Using the identity a^2-2ab+b^2=(a+b)^2 , we get x^2-2(x)(2)+c=0. Therefore, c=2^2=4.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

oh crap! i know that c = 4, but i thought of (x-2)(x-2) with x = 2. Then I clicked 2 -_-

Eka Kurniawan - 7 years, 3 months ago

Ok

Ravindra Joshi - 7 years, 2 months ago
Dominika Krawiec
May 4, 2014

For a=1and b=4 you can solve it by finding the discriminant.

b^2-4ac

16-4c=0

4c=16

c=4

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Shibsankar Das
Mar 17, 2014

16-4c=0... so c=4

0 Helpful 0 Interesting 0 Brilliant 0 Confused

given equation x^2-4x+c=0 . We have to find a value for x , so that x have only one solution. x will have only one solution if we plot a graph of the equation and it intersects same point twice. This will happen only when the given equation will represent a whole square form.

Now we will try to represent given equation into (x-p)^2=0, where p is some other value.

Now, x^2-4x+c=0 => x^2-2.2.x+c=0

we know that (a-b)^2 = a^2-2ab+b^2 so that if c=4 then, x will have one solution.

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Corner Brian
Mar 12, 2014

(a=1, b=-4, c=?) The Equation has only one solution when:

Delta = 0

While Delta = b^2 - 4ac

=> 16-4c = 0

=> c=4

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Sathi Nagi Reddy
Mar 11, 2014

find out the roots with delta formulae x=(-b±√(b^2-4ac))/2a. Then equate the roots and solve it. You will get the answer as 4

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Raghda Allam
Mar 11, 2014

a^2-4a+x=0 (a-2)(a-2)=0 this it get only answer so, x=4

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Sharat Bhat
Mar 7, 2014

x^2-4x+c=0 will have only one solution when it is a square number. Using the identity a^2-2ab+b^2=(a+b)^2, we get x^2-2(x)(2)+c=0. Therefore, c=2^2=4.

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