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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: Q. 19 (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 $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ .
$3 x^{2} + 3 x = 4$

Short Answer

Expert verified

The values of x are $x = \frac{- 3 + \sqrt{57}}{6}$ and $x = \frac{- 3 - \sqrt{57}}{6}$ .

Step by step solution

01

- Understand the concept used

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

02

- Substitute the values

It is given that $3 x^{2} + 3 x = 4$ which can also be written as $3 x^{2} + 3 x - 4 = 0$ . Using quadratic formula $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ , substitute 3 for a, 3 for b and -4 for c.

Put the values in the formula and get $x = \frac{- 3 \pm \sqrt{{(3)}^{2} - 4 \times (3) \times (- 4)}}{2 (3)}$ .

03

- Simplify for x

Simplify the expression for x.

$\begin{matrix} x = \frac{- 3 \pm \sqrt{{(3)}^{2} - 4 \times (3) \times (- 4)}}{2 (3)} \\ = \frac{- 3 \pm \sqrt{9 + 48}}{6} \\ = \frac{- 3 \pm \sqrt{57}}{6} \end{matrix}$

Therefore, $x = \frac{- 3 + \sqrt{57}}{6}$ and $x = \frac{- 3 - \sqrt{57}}{6}$ .

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Most popular questions from this chapter

Decide whether you can deduce by the SSS, SAS, or ASA postulate that another triangle is congruent to $Δ A B C$ . If so, write the congruence and name the postulate used. If not, write no congruence can be deduced.

Given: $Δ ABC$ in plane M; D not in plane M;

$∠ ACB ≅ ∠ ABC; ∠ DCB ≅ ∠ DBC$

Name a pair of congruent triangles. Prove that your answer is correct.

Plot the given points on graph paper. Draw $Δ ABC$ and $Δ DEF$ . Copy and complete the statement $Δ ABC ≅ \underline{?}$ .

$A (- 3, 1)$ $B (2, 1)$ $C (2, 3)$

$D (4, 3)$ $E (6, 3)$ $F (6, 8)$

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

Write proof in two-column form.

Given: $\bar{AD} ⊥ \bar{BC}$ ; $\bar{BA} ⊥ \bar{AC} .$

Prove: $∠ 1 ≅ ∠ 2 .$

See all solutions

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