# Equations- Factorization of a polynomial equation – Exercise 5598

Exercise

$$9x^2-42x+49=0$$

$${(3x-7)}^2$$

Solution

$$9x^2-42x+49=0$$

It’s a quadratic equation with the coefficients:

$$a=9, b=-42, c=49$$

We solve it with the quadratic formula. Putting the coefficients in the formula gives us

$$x_{1,2}=\frac{42\pm \sqrt{{42}^2-4\cdot 9\cdot 49}}{2\cdot 9}=$$

$$=\frac{42\pm 0}{18}=$$

$$=\frac{14}{6}=\frac{7}{3}$$

We got one solution for a quadratic equation, so it appears twice in the factorizing of the equation:

$$9x^2-42x+49=$$

$$=9{(x-\frac{7}{3})}^2=$$

we can get rid of the fraction:

$$=3^2{(x-\frac{7}{3})}^2=$$

$$={(3(x-\frac{7}{3}))}^2=$$

$$={(3x-7)}^2$$

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