Question

Find the value of a, b, and c of the following quadratic equations. Solve for the roots given the indicated method. In a clean sheet of paper, a. copy the picture below b. write your answer on the table c. show your solutions d. take a picture of your answer e. upload the picture or document of your answer. (WRITE YOUR NAME AND SECTION ON THE PICTURE/DOCUMENT THAT YOU WILL UPLOAD)

pa sagot po salamat❤️!​

Answer 1

Answer:

4-5 9p2- is 9p ok please make me brainlist

Answer 2

Tip:

• Quadratic equation: $ax^2+bx+c=0$.

Explanation:

• Given four quadratic equations and their method for finding the roots.
• We have find $a,b,c$ of the quadratic equation and their roots from given method.
• We will solve them with the help of given methods.

Step

Step 1 of 4:

The quadratic equation is

$9(2x-1)^2-8=0$

On further solving it, we get

$9(4x^2+1-4x)-8=0\\36x^2-36x+1=0$

So, $a=36,~b=-36,~c=1$

For roots,

By Extracting the square roots, we get

$9(2x-1)^2-8=0\\(2x-1)^2=\frac{8}{9}\\2x-1=\pm\frac{2\sqrt{2}}{3}\\x=\frac{3\pm2\sqrt{2}}{6}$

So, the roots are $\frac{3+2\sqrt{2}}{6},~\frac{3-2\sqrt{2}}{6}$

Step 2 of 4:

The quadratic equation is

$2x^2+5x=12$

On solving it,

$2x^2+5x-12=0$

So, $a=2,~b=5,~c=-12$

For roots,

By Factorization, we get

$2x^2+5x-12=0\\2x^2+8x-3x-12=0\\2x(x-4)-3(x-4)=0\\(2x-3)(x-4)=0\\x=\frac{3}{2},~4$

So, the roots are $\frac{3}{2},~4$

Step 3 of 4:

The quadratic equation is

$m^2-5m=0$

So, $a=1,~b=-5,~c=0$

For roots,

By Completing the Square Method, we get

$m^2-5m=0\\m^2-5m+\left(\frac{5}{2}\right)^2=\left(\frac{5}{2}\right)^2\\\left(m-\frac{5}{2}\right)^2=\frac{25}{4}\\\left(m-\frac{5}{2}\right)=\pm\frac{5}{2}$

$m=\frac{5}{2}\pm\frac{5}{2}\\m=\frac{5}{2}+\frac{5}{2},\frac{5}{2}-\frac{5}{2}\\m=5,0$

So, the roots are $5,~0$

Step 4 of 4:

The quadratic equation is

$9p^2-7=9p$

On further simplifying it,

$9p^2-9p-7=0$

So, $a=9,~b=-9,~c=-7$

For roots,

By Quadratic formula, we get

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

Putting the value of $a,b,c$

$p=\frac{9\pm\sqrt{81-4(9)(-7)}}{2(9)}\\p=\frac{9\pm\sqrt{81-252}}{18}\\p=\frac{9\pm\sqrt{-171}}{18}\\p=\frac{3\pm\sqrt{19}}{6}$

$p=\frac{3+\sqrt{19}}{6},~\frac{3-\sqrt{19}}{6}$

So, the roots are $\frac{3+\sqrt{19}}{6},~\frac{3-\sqrt{19}}{6}$

Final Answer:

• The equation $9(2x-1)^2-8=0$ has roots $\frac{3+2\sqrt{2}}{6},~\frac{3-2\sqrt{2}}{6}$.
• The equation $2x^2+5x=12$ has roots  $\frac{3}{2},~4$.
• The equation $m^2-5m=0$ has roots $5,~0$.

• The equation $9p^2-7=9p$ has roots $\frac{3+\sqrt{19}}{6},~\frac{3-\sqrt{19}}{6}$