funsogu wrote:
I am confused sudarshan22. Thought the roots should be 6(1+√2) and 6(1-√2) with product of -36 (
A). Not 6 and -6 even though their product is -36.
sudarshan22 wrote:
Bunuel wrote:
If \(\frac{(x - 3)^2}{(2x + 15)} = 3\), what is the product of the possible values of x?
(x - 3)^2 / (2x + 15) = 3
x^2 - 6x + 9 = 6x + 45
x^2 - 12x - 36 = 0
On solving x = 6, -6
Product = -36
Hence,
A.
funsoguyou are right
\(\frac{(x - 3)^2}{(2x + 15)} = 3\)
\((x - 3)^2=3(2x + 15)\)
\(x^2 + 9 -6x =6x +45\)
\(x^2 + 9-45 -6x-6x =0\)
\(x^2 -12x -36 =0\)
Roots \(=\frac {-b±\sqrt{b^2-4ac}}{2a}\) where \(ax^2+bx+c=0\)
\(a=1,b=-12,c=-36\)
Roots\(= \frac{12+ \sqrt{(-12)^2 - 4*1*(-36)}}{2*1}; \frac{12- \sqrt{(-12)^2 - 4*1*(-36)}}{2*1}\)
Roots\(= \frac{12+ \sqrt{2*12^2}}{2}; \frac{12- \sqrt{2*12^2}}{2}\)
Roots\(= 6(1+ \sqrt{2});6 (1- \sqrt{2})\)
It is better to follow the method as followed by
pradeep15793Quote:
Instead of finding roots and multiplying them, we can get product directly.
\(x^2-6x+9 = 6x+45\)
\(x^2-12x-36 = 0\)
Product of roots for equation \(ax^2+bx+c=0\) is \(\frac{c}{a}\)
So, product of roots\(= -36\)
Hence option A
This can save time in exam