Question

Pls someone should kindly help me with this

Answer 1

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

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

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

Answer 2

Answer: 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 = zero
• Case 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