Question

Find the value of m so that the quadratic equation mx(x – 3) + 9 = 0

has two equal roots​

Answer 1

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Answer:

Mx(x-3)+9 = 0

mx^2 - 3mx +9 = 0

mx^2 - 3mx =0-9

mx^2 - 3mx = -9

Answer 2

Answer:

The value of $m=4$ .

Step-by-step explanation:

A quadratic equation is an equation of the form $ax^{2} +bx=c=0$  where $a\neq 0$  . Since it contains the term of $x^{2}$  , $x$ has two values.

These  values can be found in three methods

1. Using quadratic formula
2. By the method of factorization
3. Completing the square method.

In this question we are asked to find the value of $m$.

Nature of roots are determined by calculating the discriminant  $b^{2} -4ac$ .

If $b^{2} -4ac>0$ , then the roots are real and distinct.

If $b^{2} -4ac=0$ , then the roots are equal .

If $b^{2} -4ac<0$ , then the equation has no root in real numbers.

Here it is given that the roots are equal .Therefore $b^{2} -4ac=0$.

Given

          [tex]mx(x-3)+9=0\\ \\ mx^{2} -3mx+9=0[/tex]

Thus $a=m,\ b=-3m,\ c=9$

$b^{2} -4ac=(-3m)^{2} -4(m)9$

             [tex]=9m^{2} -36m=0\\ \\ =m(9m-36)=0[/tex]

Therefore either $m=0$ or $m=4$

$m=0$ is not possible because it didn't satisfies the equation.

Hence the value of $m=4$ .

Thus the answer.

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