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melamori03 [73]

2 years ago

14

I need help on this one please

1 answer:

zepelin [54]2 years ago

3 0

Answer:

here is what i think

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. Use the quadratic formula to solve each quadratic real equation. Round

Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

$\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}$ , with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}$

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}$

$\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5$

So B has two solutions of 5 and -1.5.

Now to C!

$\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}$

$\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}$

$\frac{44}{8} = 5.5$

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

4 0

2 years ago

If the squared difference of the zeroes of the quadratic polynomial x2+kx+30 is equal to 169 find the value of k and the zeroes

anyanavicka [17]

ANSWER

$x = 2 \: \: or \: \: x = 15$

Or

$x = - 2 \: \: or \: \: x = - 15$

EXPLANATION

The given polynomial is

$f(x) = {x}^{2} + kx + 30$

where a=1,b=k, c=30

Let the zeroes of this polynomial be m and n.

Then the sum of roots is

$m + n = - \frac{b}{a} = -k$

and the product of roots is

$mn = \frac{c}{a} = 30$

The square difference of the zeroes is given by the expression.

$( {m - n})^{2} = {(m + n)}^{2} - 4mn$

From the question, this difference is 169.

This implies that:

$( { - k)}^{2} - 4(30) = 169$

${ k}^{2} -120= 169$

$k^{2} = 289$

$k= \pm \sqrt{289}$

$k= \pm17$

We substitute the values of k into the equation and solve for x.

$f(x) = {x}^{2} \pm17x + 30$

$f(x) = (x \pm2)(x \pm 15)$

The zeroes are given by;

$(x \pm2)(x \pm 15) = 0$

$x = \pm2 \: \: or \: \: x = \pm 15$

5 0

3 years ago

The system has infinitely many solutions. Rearranging the left side of the 2nd equation and subtracting 3a from both sides gives

Alex_Xolod [135]

Answer:

https://web.williams.edu/Mathematics/sjmiller/public_html/105/hwsolns/HWSolns_Math105_Sp2013.pdf click the link

Step-by-step explanation:

5 0

2 years ago

OMG GUYS I NEED SOME HELP I DIDNT KNOW MY SCHOOL ENDS TODAY AND IM FREAKING OUTT!!! ILL GIVE ALL MY POINTS AWAY IF SOMEONE HELPS

Anna007 [38]

What school is it, and what help do you need?

7 0

3 years ago

Read 2 more answers

Last week, Robert's Diner sold 19 milkshakes with whipped cream on top and 19 milkshakes without whipped cream. What percentage

bagirrra123 [75]

Answer:

50 percentage

Step-by-step explanation:

19+19=38

he had 19 whipped cream

cuz that's half 38

4 0

2 years ago