Question

Please Answer fast .......​

Answer 1

Answer:

  3)  67/441

Step-by-step 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

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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

Answer 2

Answer:

Option 3 $\frac{67}{441}$

Step-by-step 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

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

$x=\frac{-5(+/-)\sqrt{109}} {6}$

$x=\frac{-5+\sqrt{109}} {6}$

$x=\frac{-5-\sqrt{109}} {6}$

step 2

Let

$\alpha=\frac{-5+\sqrt{109}} {6}$

$\beta=\frac{-5-\sqrt{109}} {6}$

we need to calculate

$\frac{1}{(3\alpha+5)^{2}}+ \frac{1}{(3\beta+5)^{2}}$

step 3

Calculate   $(3\alpha+5)^{2}$

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

$=[(\frac{-5+\sqrt{109}} {2})+5]^{2}$

$=[(\frac{-5+\sqrt{109}+10} {2})]^{2}$

$=[(\frac{5+\sqrt{109}} {2})]^{2}$

$=[(\frac{25+10\sqrt{109}+109} {4})]$

$=[(\frac{134+10\sqrt{109}} {4})]$

$=[(\frac{67+5\sqrt{109}} {2})]$

step 4

Calculate   $(3\beta+5)^{2}$

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

$=[(\frac{-5-\sqrt{109}} {2})+5]^{2}$

$=[(\frac{-5-\sqrt{109}+10} {2})]^{2}$

$=[(\frac{5-\sqrt{109}} {2})]^{2}$

$=[(\frac{25-10\sqrt{109}+109} {4})]$

$=[(\frac{134-10\sqrt{109}} {4})]$

$=[(\frac{67-5\sqrt{109}} {2})]$

step 5

substitute

$\frac{1}{(3\alpha+5)^{2}}+ \frac{1}{(3\beta+5)^{2}}$

$\frac{1}{[(\frac{67+5\sqrt{109}} {2})]}+ \frac{1}{[(\frac{67-5\sqrt{109}} {2})]}$

$\frac{2}{67+5\sqrt{109}} +\frac{2}{67-5\sqrt{109}}\\ \\\frac{2(67-5\sqrt{109})+2(67+5\sqrt{109})}{(67+5\sqrt{109})(67-5\sqrt{109})} \\ \\\frac{268}{1764}$

Simplify

Divide by 4 both numerator and denominator

$\frac{268}{1764}=\frac{67}{441}$