Algebra 1 Common Core Student Edition, Grade 8-9

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Can you use the
quadratic formula to
solve part (A)?
Yes. You can use th e
quadratic formula w ith
a = 3, b = 0, and
c = - 9. However, it
is fa ste r to use square
roots.

Pr o b l em 3 Ch o o si n g a n A p p r o p r i at e M et h o d

Which method(s) would you choose to solve each equation? Explain your reasoning.

Q 3x2 — 9 = 0 Square roots; there is no x-term

Q x2 — x — 30 = 0 Factoring; the equation is easily factorable

0 6x2 + 13x — 17 = 0 Quadratic formula, graphing; the equation cannot be factored

0 x 2 — 5x + 3 = 0 Quadratic formula, completing the square, or graphing; the

coefficient of the x2-term is 1, but the equation cannot be

factored

Q — 16x2 — 50x + 21 = 0 Quadratic formula, graphing; the equation cannot be factored

easily since the numbers are large

Go t I t? 3. Which method(s) would you choose to solve each equation? Justify your

reasoning.

a. x2 - 8 x + 12 = 0 b. 169x2 = 36 c. 5x2 + 13x - 1 = 0

Quadratic equations can have two, one, or no real-number solutions. Before you solve

a quadratic equation, you can determine how many real-number solutions it has by

using the discriminant. The discrim inant is the expression under the radical sign in the

quadratic formula.

—b ± Vb2 -4ac

2 a

the discrim inant

The discriminant of a quadratic equation can be positive, zero, or negative.

|(ey Concept Usi ng t he Di scr i mi nant

Discriminant b2 -4 a c > 0 b2 -4 a c = 0 b2-4 a c< 0

Example x2 - 6x + 7 = 0

The discriminant is

(~6)2 - 4(1)(7) = 8,

which is positive.

x z — 6x + 9 = 0

The discriminant is

( 6)2 - 4(1)(9) = 0.

yi

\ / X

0 \, ■ /

\

Number of

Solutions

There are two real-

number solutions.

There is one real-

number solution.

x2 — 6x + 11 = 0

The discriminant is

( 6)2 — 4(1)(11) = -

which is negative.

8 ,

There are no real-

number solutions.

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Lesso n 9- 6 Th e Q u ad r at i c Fo r m u l a an d t h e D i scr i m i n an t^585