Question

Please Answer fast .......​

Answer 1

Option 3
`(67)/(441)`

Explanation:

step 1

Find the roots of the quadratic equation

we have

`3x^(2)+5x-7=0`

The formula to solve a quadratic equation of the form
`ax^(2) +bx+c=0` is equal to

`x=\frac{-b(+/-)\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`3x^(2)+5x-7=0`

so

`a=3\\b=5\\c=-7`

substitute in the formula

`x=\frac{-5(+/-)\sqrt{5^(2)-4(3)(-7)}} {2(3)}`

`x=\frac{-5(+/-)√(109)} {6}`

`x=\frac{-5+√(109)} {6}`

`x=\frac{-5-√(109)} {6}`

step 2

Let

`\alpha=\frac{-5+√(109)} {6}`

`\beta=\frac{-5-√(109)} {6}`

we need to calculate

`(1)/((3\alpha+5)^(2))+ (1)/((3\beta+5)^(2))`

step 3

Calculate
`(3\alpha+5)^(2)`

`(3\alpha+5)^(2)=[3(\frac{-5+√(109)} {6})+5]^(2)`

`=[(\frac{-5+√(109)} {2})+5]^(2)`

`=[(\frac{-5+√(109)+10} {2})]^(2)`

`=[(\frac{5+√(109)} {2})]^(2)`

`=[(\frac{25+10√(109)+109} {4})]`

`=[(\frac{134+10√(109)} {4})]`

`=[(\frac{67+5√(109)} {2})]`

step 4

Calculate
`(3\beta+5)^(2)`

`(3\beta+5)^(2)=[3(\frac{-5-√(109)} {6})+5]^(2)`

`=[(\frac{-5-√(109)} {2})+5]^(2)`

`=[(\frac{-5-√(109)+10} {2})]^(2)`

`=[(\frac{5-√(109)} {2})]^(2)`

`=[(\frac{25-10√(109)+109} {4})]`

`=[(\frac{134-10√(109)} {4})]`

`=[(\frac{67-5√(109)} {2})]`

step 5

substitute

`(1)/((3\alpha+5)^(2))+ (1)/((3\beta+5)^(2))`

`\frac{1}{[(\frac{67+5√(109)} {2})]}+ \frac{1}{[(\frac{67-5√(109)} {2})]}`

`(2)/(67+5√(109)) +(2)/(67-5√(109))\\ \\(2(67-5√(109))+2(67+5√(109)))/((67+5√(109))(67-5√(109))) \\ \\(268)/(1764)`

Simplify

Divide by 4 both numerator and denominator

`(268)/(1764)=(67)/(441)`

Answer 2

3) 67/441

Explanation:

Comparing the given equation to the expressions you need to evaluate, you find there might be a simplification.

3x² +5x -7 = 0 . . . . . given equation

3x² +5x = 7 . . . . . . . add 7

x(3x +5) = 7 . . . . . . . factor

3x +5 = 7/x . . . . . . . . divide by x

Now, we can substitute into the expression you are evaluating to get ...

1/(3α +5)² +1/(3β +5)² = 1/(7/α)² +1/(7/β)² = (α² +β²)/49

__

We know that when we divide the original quadratic by 3, we get

x² +(5/3)x -7/3 = 0

and that (α+β) = -5/3, the opposite of the x coefficient, and that α·β = -7/3, the constant term. The sum of squares is ...

α² +β² = (α+β)² -2αβ = (-5/3)² -2(-7/3) = 25/9 +14/3 = 67/9

Then the value of the desired expression is ...

(67/9)/49 = 67/441