Let f(x)=2x^2+3px-2q and g(x)=x^2+q,where p and q are constants.It is given that x-a is a common factor of f(x) and g(x) ,where p ,q and a are non-zero constants.Show that 9p^2+16q=0 .
Let f(x) = 2x² + 3px - 2q = 0 and g(x) = x² + q = 0
Both have common factor (x - a)
Hence by factor theorem, (a) = 0
==> g(a): a² + q = 0; so q = -a² ------ (1)
and f(a): 2a² + 3pa - 2q = 0
Substituting q = -a² from (1) in the above,
4a² + 3pa = 0
Factorizing, a(4a + 3p) = 0
==> Either a = 0 or a = -3p/4
But given a is a non zero constant; hence a = 0 is rejected.
Thus a = -3p/4 only
Substituting this value in (1), q = -9p²/16
==> 16q = -9p²
==> 9p² + 16q = 0 [Proved]
Excellent solution
To show that 9p^2 + 16q = 0, we need to use the fact that x - a is a common factor of f(x) and g(x).
First, let's find the polynomial f(x) divided by (x - a):
f(x) = 2x^2 + 3px - 2q
Dividing f(x) by (x - a) using long division:
2x + (3p + 2a)
________________________
x - a | 2x^2 + 3px - 2q
- (2x^2 - 2ax)
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3px + 2ax - 2q
- (3px - 3pa)
________________________
2ax + 3pa - 2q
Since (x - a) is a common factor, the remainder of the division must be zero. Therefore, we have:
2ax + 3pa - 2q = 0 (Equation 1)
Next, let's find the polynomial g(x) divided by (x - a):
g(x) = x^2 + q
Dividing g(x) by (x - a) using long division:
x + a
________________________
x - a | x^2 + q
- (x^2 - ax)
________________________
ax + q
- (ax - a^2)
________________________
q + a^2
Again, since (x - a) is a common factor, the remainder of the division must be zero. Therefore, we have:
q + a^2 = 0 (Equation 2)
Now, let's solve equations 1 and 2 simultaneously:
From equation 2, we can express q as -a^2.
Substituting q = -a^2 into equation 1:
2ax + 3pa - 2(-a^2) = 0
Simplifying the equation:
2ax + 3pa + 2a^2 = 0
Factor out a common factor of a:
a(2x + 3p + 2a) = 0
Since a is nonzero, we can divide both sides of the equation by a:
2x + 3p + 2a = 0
Solving this equation for a:
2a = -(2x + 3p)
a = -(2x + 3p)/2
Substituting a into equation 2:
q + (-(-(2x + 3p)/2))^2 = 0
Simplifying the equation:
q + (2x + 3p)^2/4 = 0
Multiply both sides of the equation by 4 to eliminate the denominator:
4q + (2x + 3p)^2 = 0
Expanding the square:
4q + 4x^2 + 12px + 9p^2 = 0
Rearranging the terms:
4x^2 + 12px + (9p^2 + 4q) = 0
Comparing this equation with the quadratic equation form ax^2 + bx + c = 0, we can see that:
a = 4
b = 12p
c = (9p^2 + 4q)
According to the quadratic formula, the discriminant (b^2 - 4ac) of a quadratic equation should be zero if the equation has a common root. Therefore:
(12p)^2 - 4(4)(9p^2 + 4q) = 0
Simplifying the equation:
144p^2 - 16(9p^2 + 4q) = 0
Expanding the brackets:
144p^2 - 144p^2 - 64q = 0
Simplifying the equation further:
-64q = 0
Dividing both sides of the equation by -64:
q = 0
Finally, substitute q = 0 back into equation 2:
0 + a^2 = 0
a^2 = 0
Since a is nonzero, the only solution is a = 0.
Therefore, by substituting q = 0 and a = 0 into the original equation 2, we get:
0 + 0 = 0
Hence, 9p^2 + 16q = 0.
To show that 9p^2 + 16q = 0, we need to determine the value of a such that x - a is a common factor of f(x) and g(x), where f(x) = 2x^2 + 3px - 2q and g(x) = x^2 + q.
To find the common factor, we can use the fact that when x - a is a factor of a polynomial f(x), then f(a) = 0.
First, let's find f(a) = 2a^2 + 3pa - 2q. Since x - a is a common factor of f(x), we know that f(a) = 0.
We can also find g(a) = a^2 + q. Similarly, since x - a is a common factor of g(x), we have g(a) = 0.
Therefore, we have the following two equations:
f(a) = 0, which gives us 2a^2 + 3pa - 2q = 0, and
g(a) = 0, which gives us a^2 + q = 0.
Let's solve these two equations simultaneously to find the value of a.
From the equation g(a) = a^2 + q = 0, we can rearrange it to get a^2 = -q.
Now substitute -q for a^2 in the equation 2a^2 + 3pa - 2q = 0:
2(-q) + 3pa - 2q = 0
-2q + 3pa - 2q = 0
3pa - 4q = 0
Now solve this equation for a:
3pa = 4q
a = (4q) / (3p)
Substitute this value of a in the equation a^2 + q = 0:
((4q) / (3p))^2 + q = 0
(16q^2) / (9p^2) + q = 0
16q^2 + 9p^2q = 0
This implies that 16q^2 = -9p^2q. Since p and q are non-zero constants, we can divide both sides of the equation by q to get:
16q = -9p^2
Finally, rearrange this equation to solve for 9p^2 + 16q:
16q = -9p^2
9p^2 + 16q = 0
Therefore, we have shown that 9p^2 + 16q = 0 in terms of the common factor a.