Mathematics, 27.06.2019 03:00, aleilyg2005

Pls someone should kindly me with this

Answers: 2

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Answers

answered: parkerpark9524

a = 1, a = ${\bold{-\dfrac{119}{125}}$

Step-by-step explanation:

In order to have the same root, the discriminant cannot be irrational.

Case 1: Discriminant = zeroCase 2: Discriminant = perfect square

(a + 3)x² - (11a + 1)x + a = 2(a - 5)

(a + 3)x² - (11a + 1)x + a = 2a - 10

(a + 3)x² - (11a + 1)x + a - 2a + 10 = 0

(a + 3)x² - (11a + 1)x - (a - 10) = 0

a = a+3 b = -(11a+1) c = -(a - 10)

Case 1:

b² - 4ac = 0

[-(11a + 1)]² - 4(a + 3)[-(a - 10)] = 0

121a² + 22a + 1 + 4a² - 28a - 120 = 0

125a² - 6a - 119 = 0

Use any method to solve the quadratic equation. I chose to use the factoring method.

125a² - 6a - 119 = 0

125a² - 125a + 119a - 119 = 0

125a(a - 1) + 119(a - 1) = 0

(125a + 119)(a - 1) = 0

125a + 119 = 0 and a - 1 = 0

a = $-\dfrac{119}{125}$ and a = 1

Check:

(a + 3)x² - (11a + 1)x + a = 2(a - 5)

((1) + 3)x² - (11(1) + 1)x + (1) = 2((1) - 5)

4x² - 12x + 1 = -8

4x² - 12x + 9 = 0

(2x + 3)² = 0

x = $-\dfrac{3}{2}$

Case 2: I am not sure how to do this one

answered: cmflores3245

$ax^2+bx+c=0\\\\if\ \Delta=b^2-4ac0\ then\ two\ solutions\qquad x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\if\ \Delta=b^2-4ac=0\ then\ one\ solution\\if\ \Delta=b^2-4ac$

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$(a+3)x^2-(11a+1)x+a=2(a-5)\qquad\text{use distributive property}\\\\(a+3)x^2-(11a+1)x+a=2a-10\qquad\text{subtract 2 from both sides and add 10 to both sides}\\\\(a+3)x^2-(11a+1)x-a+10=0\\\\\Delta=[-(11a+1)]^2-4(a+3)(-a+10)\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\\Delta=(11a)^2+2(11a)(1)+1^2+(-4a-12)(-a+10)\\\\\Delta=121a^2+22a+1+4a^2-40a+12a-120\\\\\Delta=125a^2-6a-119$

$\text{One solution if}\ \Delta=0.\\\\\Delta=0\iff125a^2-6a-119=0\\\\\Delta_a=(-6)^2-4(125)(-119)=36+59,500=59,536\\\\\sqrt{\Delta_a}=\sqrt{59,536}=244\\\\a_1=\dfrac{-(-6)-244}{2(125)}=\dfrac{-238}{250}=\dfrac{-238:2}{250:2}=-\dfrac{119}{125}\\\\a_2=\dfrac{-(-6)+244}{2(125)}=\dfrac{250}{250}=1\\\\\ \boxed{a=-\dfrac{119}{125}\ or\ a=1}$

answered: Guest

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

$ab^2+ac^2=bc^2$