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=(-b-\sqrt\Delta)/(2a)\ and\ x_2=(-b+\sqrt\Delta)/(2a)\\if\ \Delta=b^2-4ac=0\ then\ one\ solution\\if\ \Delta=b^2-4ac<0\ then\ no\ real\ solution`

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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\\\\√(\Delta_a)=√(59,536)=244\\\\a_1=(-(-6)-244)/(2(125))=(-238)/(250)=(-238:2)/(250:2)=-(119)/(125)\\\\a_2=(-(-6)+244)/(2(125))=(250)/(250)=1\\\\Answer:\ \boxed{a=-(119)/(125)\ or\ a=1}`

Answer 2

Answer: a = 1, a =
`{\bold{-(119)/(125)}`

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 =
`-(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 =
`-(3)/(2)`

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