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Bunuel
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Using the formula b^2 = 4 * a * c, we have,

(K+1)^2 = 4(K+4)
K^2 + 1 + 2K = 4K + 4
K^2 - 2K - 3 = 0
K = 3,-1

No option given, Bunuel, please check the options.

Checked. The question is correct. Please try again.



Sorry, my bad...

(K+1)^2 = 4(K+4)
K^2 + 1 + 2K = 4K + 16
K^2 - 2K - 15 = 0
K = 5, -3

Option D, IMO.
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Could someone please clarify this question more detailed?


(K+1)^2 = 4(K+4) --> I get this
K^2 + 1 + 2K = 4K + 16 (But where does the K^2, the 1, the 1 and the 2k, come from? I cannot figure this out.
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Could someone please clarify this question more detailed?


(K+1)^2 = 4(K+4) --> I get this
K^2 + 1 + 2K = 4K + 16 (But where does the K^2, the 1, the 1 and the 2k, come from? I cannot figure this out.

(K+1)^2 = (K+1)(K+1)= K^2+K+K+1=K^2+2K+1

Hope it helps
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Bunuel
What is the values of K for which the quadratic equation \((K + 4)x^2 + (K + 1)x + 1 = 0\) has equal roots ?

A. -3 or -5
B. -3 or -1
C. -3 or 1
D. -3 or 5
E. 3 or 5


If the question deals with equal roots we can apply the below principle :
the Discriminant has to be equal to zero , since the roots are same .

thus for an equation ax^2 +bx +c , the Discriminant is given by b^2 - 4ac
thus comparing the original condition with the discriminant , we get b = K+1 , a = K+4 , c =1
therefore , (K+1)^2 - 4*(K+4)*1=0
K^2+2K+1 -4K-16=0
k^2 -2K-15=0
K^2 +3K -5K -15 =0 or (K+3)(K-5)=0 , thus K = -3 or 5 IMO ans D
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Bunuel
What is the values of K for which the quadratic equation \((K + 4)x^2 + (K + 1)x + 1 = 0\) has equal roots ?

A. -3 or -5
B. -3 or -1
C. -3 or 1
D. -3 or 5
E. 3 or 5
­
The equation has equal roots; therefore discriminant = 0.

→ \(b^2 - 4ac = 0\)
→ \((K+1)^2 - 4.(K+1).1 = 0\)
Expand to get:
→ \((K-5)(K+3) = 0\)
→ \(K = 5, -3\)
 ­
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