SETTING THE SCENE

Consider the quadratic equation

\(x^2+6x+9=something\)

where the “something” is a number yet to be specified. The square method shows that the left side of this question is actually a perfect square. The equation is really a level 2 problem in disguise.

\(\left(x+3\right)^2=something\).

And we’ve seen before that such equations can have with 2, 1, or 0 solutions. For instance,

\(\left(x+3\right)^2=4\) has precisely two solutions since there are two roots of the number \(4\),

\(\left(x+3\right)^2=0\) has precisely one solution since there is only one square root of zero,

\(\left(x+3\right)^2=-5\) has no solutions since there are no square roots of a negative value.

So this means, going back to the original form of the equation,

\(x^2+6x+9=4\)

\(x^2+6x+9=0\)

\(x^2+6x+9=-5\)

have, respectively, 2, 1, and 0 solutions.

PRACTICE 1: Use the quadratic formula to solve each of the equations shown. Ascertain the feature of the formula that determines whether the count of solutions is going to be 2, 1, or 0.

\(x^2+6x+9=4\)

\(x^2+6x+9=0\)

\(x^2+6x+9=-5\)

READ MORE HERE: QUADRATICS PD Essay 2.2

(See too Edfinity.com/XXX for a robust source of curriculum practice problems for you collate, organise, and use.)

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