Bunuel wrote:
If \(4x^2 + 4px + 4 –3p = 0\) has two distinct roots, then which of the following must be true?
A. \(p < -4\) or \(p > 1\)
B. \(-1 < p < 4\)
C. \(p < -1\) or \(p > 4\)
D. \(–4 < p < 1\)
E. \(p > 1\)
Are You Up For the Challenge: 700 Level QuestionsI. Discriminant=D=\(b^2-4ac\)
D=0, real and same roots
D>0, real and distinct roots
D<0, complex roots
Here D=\((4p)^2-4*4*(4-3p)>0……….16p^2-16(4-3p)>0……….p^2-4+3p>0\)
\((p-1)(p+4)>0\)
This means p is outside the range from -4 to 1, both inclusive.
A
But if you did not know the Discriminant formula, we could work on \((a+b)^2=0\), where both roots are similar.
II.
Let us write in the form of \((x-a)^2\), where the roots are same.
\(4x^2 + 4px + 4 –3p = 0\)
\((2x)^2 + 2*2x*p +( \sqrt{4 –3p} )^2= 0\)
Now this is in the form \(x^2+2xy+y^2=(x+y)^2=0\)
So \(p=\sqrt{4-3p}\)
Square both sides,
\(p^2=4-3p…….p^2+3p-4=0……(p-1)(p+4)=0\)
Thus, whenever p is 1 or -4, we will have both roots similar. There may be some other conditions where the roots are not real, but let us see which option does not contain 1 or -4.
A. \(p < -4\) or \(p > 1\)……..Does not contain both -4 or 1
B. \(-1 < p < 4\) …….. contains 1
C. \(p < -1\) or \(p > 4\) …….. contains -4
D. \(–4 < p < 1\)……..Does not contain both -4 or 1
But when you substitute p=0, we get \(4x^2+4x*0+4-3*0=0…….4x^2=-4\)…..roots are not real
E. \(p > 1\) ……..Does not contain both -4 or 1, but does not address range closer to -4.
Only A possible