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Sagot :
Weare given the following quadratic equation, and asked to find all its real solutions:
9 z^2 - 30 z + 26 = 1
we subtract "1" from both sides in order to be able to use the quadratic formula if needed:
9 z^2 - 30 z + 26 - 1 = 0
9 z^2 - 30 z + 25 = 0
we notice that the first term is a perfect square:
9 z^2 = (3 z)^2
and that the last term is also a perfect square:
25 = 5^2
then we suspect that we are in the presence of the perfect square of a binomial of the form:
(3 z - 5)^2 = (3z)^2 - 2 * 15 z + 5^2 = 9 z^2 - 30 z + 25
which corroborates the factorization of the trinomial we had.
Then we have:
(3 z - 5)^2 = 0
and the only way such square gives zero, is if the binomial (3 z - 5) is zero itself, which means:
3 z - 5 = 0 then 3 z = 5 and solving for z: z = 5/ 3
Then the only real solution for this equation is the value:
z = 5/3
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