Question

example 1try it! recognize solutions of quadratic equations1. what are the solutions of each equation?a. $x^2 - 36 = 0$b. $x^2 + 6x + 9 = 0$try it! solve quadratic equations using tables2. find the solutions for $4x^2 + 3x - 7 = 0$ using a table. if approximating, give the answer to the nearest tenth.

Explanation:

Step1: Isolate $x^2$ term

$x^2 = 36$

Step2: Take square root of both sides

$x = \pm\sqrt{36} = \pm6$

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Step1: Factor the quadratic

$(x+3)^2 = 0$

Step2: Solve for $x$

$x+3=0 \implies x=-3$

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Step1: Evaluate $f(x)=4x^2+3x-7$ at integer values

$x$	$f(x)$
-1	$4(-1)^2+3(-1)-7=4-3-7=-6$
1	$4(1)^2+3(1)-7=4+3-7=0$

Step2: Identify exact and approximate roots

• Exact root at $x=1$
• Approximate root between $x=-2$ and $x=-1$:

Test $x=-1.4$: $4(-1.4)^2+3(-1.4)-7=7.84-4.2-7=-3.36$
Test $x=-1.3$: $4(-1.3)^2+3(-1.3)-7=6.76-3.9-7=-4.14$
Test $x=-1.2$: $4(-1.2)^2+3(-1.2)-7=5.76-3.6-7=-4.84$
Test $x=-1.1$: $4(-1.1)^2+3(-1.1)-7=4.84-3.3-7=-5.46$
Test $x=-1.5$: $4(-1.5)^2+3(-1.5)-7=9-4.5-7=-2.5$
The value of $f(x)$ crosses 0 closer to $x=-1.3$ (since $f(-1.3)=-4.14$, $f(-1.2)=-4.84$, $f(-1.5)=-2.5$, $f(-2)=3$; the root is $\approx -1.3$ when rounded to nearest tenth)

Answer:

1. a. $x=6$ and $x=-6$
2. b. $x=-3$ (repeated root)
3. $x=1$ and $x\approx-1.3$