In this exercise, we want to know the x-intercepts of each item. To find the x-intercepts, set y = 0 as indicated in each item and solve for x. So:

<h2>1. Answer:</h2>

B. x=-1; x=-1.75

<h3>Step by step explanation:</h3>

we have the equation:

$4x^2+11x+7=0$

We can say that this equation comes from the function $f(x)=4x^2+11x+7$ so we have set $y=0$ to find the x-intercepts. By using the quadratic formula we have:

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ where: \\ \\ a=4, \ b=11, \ c=7 \\ \\ x=\frac{-11 \pm \sqrt{11^2-4(4)(7)}}{2(4)} \\ \\ x=\frac{-11 \pm \sqrt{121-112}}{8} \\ \\ \boxed{x_{1}=-1 \ and \ x_{2}=-1.75}$

<h2>2. Answer:</h2>

B. x=-1; x=-1.75

<h3>Step by step explanation:</h3>

we have the equation:

$3x^2-4x+1=0$

We can establish a function $g(x)=3x^2-4x+1$ and say that we want to find the x-intercepts of this function by setting y = 0. Therefore, by using the quadratic formula we have:

$a=3, \ b=-4, \ c=1 \\ \\ x=\frac{-(-4) \pm \sqrt{(-4)^2-4(3)(1)}}{2(3)} \\ \\ x=\frac{4 \pm \sqrt{16-12}}{6} \\ \\ \boxed{x_{1}=1 \ and \ x_{2}=\frac{1}{3}}$

<h2>3. Answer:</h2>

H. No Solution

<h3>Step by step explanation:</h3>

we have the equation:

$3x^2-4x+2=0$

We can establish a function $h(x)=3x^2-4x+2$ and say that we want to find the x-intercepts of this function by setting y = 0. Therefore, by using the quadratic formula we have:

$a=3, \ b=-4, \ c=2 \\ \\ x=\frac{-(-4) \pm \sqrt{(-4)^2-4(3)(2)}}{2(3)} \\ \\ x=\frac{4 \pm \sqrt{16-24}}{6}$

Since 16 - 24 = -8, that is, a number less than zero which is within a square root, we say that the equation $3x^2-4x+2=0$ has no any real solution.

<h2>4. Answer:</h2>

E. x=1

<h3>Step by step explanation:</h3>

we have the equation:

$x^2-2x+1=0$

We can establish a function $c(x)=x^2-2x+1$ . By setting y = 0 we'll find the x-intercepts. Let's solve this problem using other method. You can find some binomial products having a special form. So it's easier to find a solution by using distributive. The form of this polynomial is a Square of a Binomial in the form:

$(x-1)^2=0 \\ \\ Because: \\ \\ (x-1)^2=(x-1)(x-1)=x^2-x-x+1= x^2-2x+1$

Therefore, the value that satisfies this equation is $\boxed{x=1}$

<h2>5. Answer:</h2>

K. x = -1

<h3>Step by step explanation:</h3>

we have the equation:

$x^2+2x+1=0$

We can establish a function $a(x)=x^2+2x+1$ . By setting y = 0 we'll find the x-intercepts. We are going to solve this problem by using the previous method. The form of this Square of a Binomial is:

$(x+1)^2=0 \\ \\ Because: \\ \\ (x+1)^2=(x+1)(x+1)=x^2+x+x+1= x^2+2x+1$

Therefore, the value that satisfies this equation is $\boxed{x=-1}$

<h2>6. Answer:</h2>

N) x = 1/2

<h3>Step by step explanation:</h3>

we have the equation:

$4x^2-4x+1=0$

We can establish a function $b(x)=4x^2-4x+1$ and say that we want to find the x-intercepts of this function by setting y = 0. Here we will use the quadratic formula, so:

$a=4, \ b=-4, \ c=1 \\ \\ x=\frac{-(-4) \pm \sqrt{(-4)^2-4(4)(1)}}{2(4)} \\ \\ x=\frac{4 \pm \sqrt{16-16}}{8} \\ \\ \boxed{x=\frac{1}{2}}$

So we have just one solution.

<h2>7. Answer:</h2>

M) x = -1/2

<h3>Step by step explanation:</h3>

we have the equation:

$4x^2+4x+1=0$

We can establish a function $b(x)=4x^2+4x+1$ and say that we want to find the x-intercepts of this function by setting y = 0. As in the previous exercise, we will use the quadratic formula, so:

$a=4, \ b=4, \ c=1 \\ \\ x=\frac{-4 \pm \sqrt{(4)^2-4(4)(1)}}{2(4)} \\ \\ x=\frac{-4 \pm \sqrt{16-16}}{8} \\ \\ \boxed{x=-\frac{1}{2}}$

So we have just one solution.

<h2>8. Answer:</h2>

D) x = -1.45; x=1.25

<h3>Step by step explanation:</h3>

we have the equation:

$5x^2+x-9=0$

We can establish a function $D(x)=5x^2+x-9$ and say that we want to find the x-intercepts of this function by setting y = 0. By using the quadratic formula we can solve this problem, so:

$a=5, \ b=1, \ c=-9 \\ \\ x=\frac{-1 \pm \sqrt{(1)^2-4(5)(-9)}}{2(5)} \\ \\ x=\frac{-1 \pm \sqrt{1+180}}{10} \\ \\ \boxed{x_{1}=-1.45 \ and \ x_{2}=1.25}$

<h2>9. Answer:</h2>

J) x = 4; x=-3

<h3>Step by step explanation:</h3>

we have the equation:

$-x^2+x+12=0$

We can establish a function $k(x)=-(x^2-x-12)$ and say that we want to find the x-intercepts of this function by setting y = 0. In this exercise we'll use other method. Since this is a non-perfect square trinomial, we know that: