If m-nx+28x²+12x³+9x⁴ is a perfect square, then find the values of m and n.
Answers
Explanation:
It is given that the polynomial m−nx+28x
2
+12x
3
+9x
4
is a perfect square, we must equate it to the square of general form of equation that is (ax
2
+bx+c)
2
as shown below:
9x
4
+12x
3
+28x
2
−nx+m=(ax
2
+bx+c)
2
⇒9x
4
+12x
3
+28x
2
−nx+m=(ax
2
)
2
+(bx)
2
+(c)
2
+(2×ax
2
×bx)+(2×bx×c)+(2×c×ax
2
)
(∵(a+b+c)
2
=a
2
+b
2
+c
2
+2ab+2bc+2ca)
⇒9x
4
+12x
3
+28x
2
−nx+m=a
2
x
4
+b
2
x
2
+c
2
+2abx
3
+2bcx+2acx
2
Now, comparing the coefficients, we get:
a
2
=9,b
2
+2ac=28,c
2
=m,2ab=12,2bc=−n
a
2
=9
⇒a=3
2ab=12
⇒2×3×b=12
⇒6b=12
⇒b=2
b
2
+2ac=28
⇒2
2
+(2×3c)=28
⇒4+6c=28
⇒6c=28−4
⇒6c=24
⇒c=4
c
2
=m
⇒m=4
2
⇒m=16
2bc=−n
⇒n=−2bc
⇒n=−2×2×4
⇒n=−16
Hence, m=16 and n=−16.
Answer:
m=16 ,n=-16
Explanation:
It is given that polynomial m-nx+28x^2+12x^3+9x^4 is a perfect square ,we must equate it to the square of general form of equation that is (ax^2+bx+c)^2 as shown below :
9x^4+12x^3+28x^2-nx+m=(ax^2+bx+c)^2
9x^4+12x^3+28x^2-nx+m=(ax^2)^2+(bx) ^2+(c)^2
+(2ax^2*bx)+(2*bx*c)
+(2*c*ax^2)
[since (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac)
9x^4+12x^3+28x^2-nx+m=a^2x^4+b^2x^2+c^2
+2abx^3+2bcx+2acx^2
Now, comparing the coefficients,we get
a^2=9
b^2+2ac=28
c^2=m
2ab=12
2bc= -n
a^2=9
a=3
2ab=12
2*3*b=12
6b=12
b=2
b^2+2ac=28
(2)^2+2*3*c=28
4+6c=28
6c=28-4
6c=24
c=4
c^2=m
m=4^2
m=16
2bc= -n
n= -2bc
n= -2*2*4
n= -16
So, m=16
n= -16