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3 years ago

I need help from question 11- 16! Please help!

Mathematics

1 answer:

insens350 [35]3 years ago

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<h2>11. Find discriminant.</h2>

Answer: D) 0, one real solution

A quadratic function is given of the form:

$ax^2+bx+c=$

We can find the roots of this equation using the quadratic formula:

$x_{12}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Where $\Delta=b^2-4ac$ is named the discriminant. This gives us information about the roots without computing them. So, arranging our equation we have:

$4a^2-4a-6=-7 \\ \\ Adding \ 7 \ to \ both \ sides \ of \ the \ equation: \\ \\ 4a^2-4a-6+7=-7+7 \\ \\ 4a^2-4a+1=0 \\ \\ Then \ the \ discriminant: \\ \\ \Delta=(-4)^2-4(4)(1) \\ \\ \Delta=16-16 \\ \\ \boxed{Delta=0}$

<em>Since the discriminant equals zero, then we just have one real solution.</em>

<h2>12. Find discriminant.</h2>

Answer: D) -220, no real solution

In this exercise, we have the following equation:

$-r^2-2r+14=-8r^2+6$

So we need to arrange this equation in the form:

$ax^2+bx+c=$

Thus:

$-r^2-2r+14=-8r^2+6 \\ \\ Adding \ 8r^2 \ to \ both \ sides \ of \ the \ equation: \\ \\ -r^2-2r+14+8r^2=-8r^2+6+8r^2 \\ \\ Associative \ Property: \\ \\ (-r^2+8r^2)-2r+14=(-8r^2+8r^2)+6 \\ \\ 7r^2-2r+14=6 \\ \\ Subtracting \ 6 \ from \ both \ sides: \\ \\ 7r^2-2r+14-6=6-6 \\ \\ 7r^2-2r+8=0$

So the discriminant is:

$\Delta=(-2)^2-4(7)(8) \\ \\ \Delta=4-224 \\ \\ \boxed{\Delta=-220}$

<em>Since the discriminant is less than one, then there is no any real solution</em>

<h2>13. Value that completes the squares</h2>

Answer: C) 144

What we need to find is the value of $c$ such that:

$x^2+24x+c=0$

is a perfect square trinomial, that are given of the form:

$a^2x^2\pm 2axb+b^2$

and can be expressed in squared-binomial form as:

$(ax\pm b)^2$

So we can write our quadratic equation as follows:

$x^2+2(12)x+c \\ \\ So: \\ \\ a=1 \\ \\ b=12 \\ \\ c=b^2 \therefore c=12^2 \therefore \boxed{c=144}$

Finally, the value of $c$ that completes the square is 144 because:

$x^2+24x+144=(x+12)^2$

<h2>14. Value that completes the square.</h2>

Answer: C) $\frac{121}{4}$

What we need to find is the value of $c$ such that:

$z^2+11z+c=0$

So we can write our quadratic equation as follows:

$z^2+2\frac{11}{2}z+c \\ \\ So: \\ \\ a=1 \\ \\ b=\frac{11}{2} \\ \\ c=b^2 \therefore c=\left(\frac{11}{2}\left)^2 \therefore \boxed{c=\frac{121}{4}}$

Finally, the value of $c$ that completes the square is $\frac{121}{4}$ because:

$z^2+11z+\frac{121}{4}=(x+\frac{11}{2})^2$

<h2 /><h2>15. Rectangle.</h2>

In this problem, we need to find the length and width of a rectangle. We are given the area of the rectangle, which is 45 square inches. We know that the formula of the area of a rectangle is:

$A=L\times W$

From the statement we know that the length of the rectangle is is one inch less than twice the width, this can be written as:

$L=2W-1$

So we can introduce this into the equation of the area, hence:

$A=L\times W \\ \\ \\ Where: \\ \\ W:Width \\ \\ L:Length$

$A=(2W-1)(W) \\ \\ But \ A=45: \\ \\ 45=(2W-1)(W) \\ \\ Distributive \ Property:\\ \\ 45=2W^2-W \\ \\ 2W^2-W-45=0 \\ \\ Quadratic \ Formula: \\ \\ x_{12}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\ W_{1}=\frac{-(-1)+ \sqrt{(-1)^2-4(2)(-45)}}{2(2)} \\ \\ W_{1}=\frac{1+ \sqrt{1+360}}{4} \therefore W_{1}=5 \\ \\ W_{2}=\frac{-(-1)- \sqrt{(-1)^2-4(2)(-45)}}{2(2)} \\ \\ W_{2}=\frac{1- \sqrt{1+360}}{4} \therefore W_{2}=-\frac{9}{2}$