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Geometry

Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

$Math Studyset Vaia Explanations$ Math

Student Edition · 227 pages

ISBN: 9780395977279

Chapter 4: Q11. (page 163)

Solve each equation by factoring or by using the quadratic formula. The quadratic formula is:
If $a x^{2} + b x + c = 0$ , with $a \neq 0$ , then $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ .
$y^{2} - 6 y + 9 = 0$

Short Answer

Expert verified

The values of y are $y = 3$ and $y = 3$ .

Step by step solution

01

- Understand the concept used

If $a x^{2} + b x + c = 0$ , with $a \neq 0$ , then $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ . Solving using factoring can be done by splitting the middle term of equation and then making groups and equating them to zero.

02

- Substitute the values

It is given that $y^{2} - 6 y + 9 = 0$ .Using quadratic formula $y = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ , substitute 1 for a, -6 for b and 9 for c.

Put the values in the formula and get $y = \frac{- (- 6) \pm \sqrt{{(- 6)}^{2} - 4 \times (1) \times (9)}}{2 (1)}$ .

03

- Simplify for y

Simplify the expression for y.

$\begin{matrix} y = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = \frac{- (- 6) \pm \sqrt{{(- 6)}^{2} - 4 (1) (9)}}{2 (1)} \\ = \frac{6 \pm \sqrt{36 - 36}}{2} \\ = \frac{6 \pm 0}{2} \end{matrix}$

Therefore, either $y = \frac{6 + 0}{2} = \frac{6}{2}$ or $y = \frac{6 - 0}{2} = \frac{6}{2}$ that is., either $y = 3$ or $y = 3$ .

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Most popular questions from this chapter

For the following figure, does the SAS postulates justify that the two triangles are congruent.

Suppose you are given a scalene triangle and a point $P$ on some line $l$ . How many triangles are there with one vertex at $P$ , another vertex on $l$ , and each triangle congruent to the given triangle.

In the following figure, the two-triangle shown are congruent. Then complete the following statement.

$B O =___$

In the following figure, the two-triangle shown are congruent. Then complete the following statement.

$Δ A B O ≅ Δ___$

For the following figure, can the triangle be proved congruent. If so, what postulate can be used?

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