Question

If (x-a) is the factor of 3x²-mx-na, then prove that a=m+n/3​

Answer 1

Step-by-step explanation :

Given :

(x - a) is the factor of 3x²-mx-na

To prove :

 a = (m+n)/3

Solution :

Given polynomial, 3x²-mx-na

Let p(x) = 3x²-mx-na

⇒ (x - a) is a factor

x - a = 0

 x = a

Since it is a factor, when we substitute x = a in the polynomial, the result should be zero.

p(a) = 0

3(a)²- m(a) - na = 0

3(a²) - ma - na = 0

3a² - ma - na = 0

a(3a - m - n) = 0

 3a - m - n = 0/a

 3a - m - n = 0

 3a = m + n

  a = (m + n)/3

Hence proved!

______________________

Know more :

• ax² + bx + c is the general form of the quadratic polynomial.

      $\sf x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$    is the quadratic formula.

• Nature of roots is determined by the value of the discriminant.

⇒ D = b² - 4ac

If D > 0 ; the roots are real and unequal

If D = 0 ; the roots are real and equal

If D < 0 ; the roots are not real i.e., complex roots

Answer 2

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Solution:-

As they given (x-a) is a factor of 3x²-mx-na

So, x- a is a factor

x - a = 0

x = a

Substuite value of x in given equation

3(a²)-m(a)-na = 0

3a² -am -na = 0

a(3a -m - n) = 0

3a - m - n =0

3a = m +n

a = m+ n/3 Hence proved

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