Equations- Factorization of a polynomial equation – Exercise 5598

Exercise

Factor the quadratic equation

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

Final Answer

Show final answer

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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