Question

5-90.
solve the equations below for x. check your solutions. homework help
a. $x^{2}+6x - 40 = 0$
b. $2x^{2}+13x - 24 = 0$
5-91.
calculate the value of each expression below. then compare your answers to those in problem 5-90. what do you notice? homework help
a. $\frac{-6+sqrt{6^{2}-(4)(1)(-40)}}{2 \cdot 1}$

Explanation:

Problem 5-90a

Step1: Identify quadratic coefficients

For $x^2 + 6x - 40 = 0$, $a=1$, $b=6$, $c=-40$

Step2: Apply quadratic formula

$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$x = \frac{-6 \pm \sqrt{6^2-4(1)(-40)}}{2(1)}$$

Step3: Compute discriminant

$\sqrt{36 + 160} = \sqrt{196} = 14$

Step4: Solve for x values

$x_1 = \frac{-6 + 14}{2} = 4$, $x_2 = \frac{-6 - 14}{2} = -10$

Step5: Verify solutions

For $x=4$: $4^2 +6(4)-40=16+24-40=0$
For $x=-10$: $(-10)^2 +6(-10)-40=100-60-40=0$

Problem 5-90b

Step1: Identify quadratic coefficients

For $2x^2 +13x -24=0$, $a=2$, $b=13$, $c=-24$

Step2: Apply quadratic formula

$$x = \frac{-13 \pm \sqrt{13^2-4(2)(-24)}}{2(2)}$$

Step3: Compute discriminant

$\sqrt{169 + 192} = \sqrt{361} = 19$

Step4: Solve for x values

$x_1 = \frac{-13 + 19}{4} = \frac{6}{4} = \frac{3}{2}$, $x_2 = \frac{-13 -19}{4} = \frac{-32}{4} = -8$

Step5: Verify solutions

For $x=\frac{3}{2}$: $2(\frac{3}{2})^2 +13(\frac{3}{2})-24=2(\frac{9}{4})+\frac{39}{2}-24=\frac{9}{2}+\frac{39}{2}-24=\frac{48}{2}-24=24-24=0$
For $x=-8$: $2(-8)^2 +13(-8)-24=128-104-24=0$

Problem 5-91a

Step1: Simplify discriminant first

$\sqrt{6^2-(4)(1)(-40)} = \sqrt{36+160} = \sqrt{196}=14$

Step2: Compute numerator

$-6 + 14 = 8$

Step3: Compute denominator

$2 \cdot 1 = 2$

Step4: Divide to find value

$\frac{8}{2} = 4$

Step5: Compare to 5-90a

This equals the positive root from 5-90a.

Answer:

5-90

a. $x=4$ and $x=-10$
b. $x=\frac{3}{2}$ and $x=-8$

5-91

a. $4$; This value is one of the solutions to the quadratic equation in 5-90a (the positive root calculated via the quadratic formula).