Question

how to do these step by step? atleast do 6 for understanding.​

Answer 1

Solution 1 :

Given equation :

• y² - 4y - 1 = 0

To Find :

• Roots of the equation.

Solution :

$\mathtt{y^2-4y-1=0}$

Compare the equation with general form,

• ax² + bx + c = 0

We get values as :

• a = 1
• b = - 4
• c = - 1

Find the discrimant (Δ)

Δ = $\mathtt{b^2-4ac}$

Δ = $\mathtt{(-4)^2\:-\:4(1)(-1)}$

Δ = $\mathtt{16\:-\:4(-1)}$

Δ = $\mathtt{16\:-\:(-4)}$

Δ = $\mathtt{16+4}$

Δ = $\mathtt{20}$

Now, we have the value of the discrimant i.e b² - 4ac (Δ).

So, now using the quadratic formula.

y = -b ± (√b²-4ac) /2a

y = - (-4) ± (√20)/2(1)

y = 4 ± √(2 × 2 × 5) /2

y = 4 ± 2√5/2

y = $\mathtt{\dfrac{\pm\:2\:\sqrt{5}\:+\:4}{2}}$

y = $\mathtt{\dfrac{\pm\:2(\sqrt{5}\:+\:2)}{2}}$

y = $\mathtt{\pm\:\:2\:\:\sqrt{5}}$

Since, we can see that we have ± sign in the value of y.

So, we will have one positive and one negative root.

y = $\mathtt{\:2\:+\:\sqrt{5}}$ OR $\mathtt{2\:-\:\sqrt{5}}$

Solution 5 :

Given equation :

• 2y² + 5y - 3 = 0

To Find :

• Roots of the equation.

Solution :

$\mathtt{2y^2\:+\:5y\:-\:3=0}$

$\mathtt{2y^2\:+\:6y-y\:-3=0}$

$\mathtt{2y(y+3)-1(y+3)=0}$

$\mathtt{(y+3)\:\:(2y-1) =0}$

$\mathtt{y+3=0\:\:or\:\:2y-1=0}$

$\mathtt{y=-3\:\:or\:\:2y=1}$

$\mathtt{y=-3\:\:or\:\:y\:=\dfrac{1}{2}}$

Solution 6 :

Given equation :

• -2y² + y + 1 = 0

To Find :

• Roots of the equation.

Solution :

Divide throughout by minus,

$\mathtt{2y^2-y-1=0}$

$\mathtt{2y^2\:-2y+y-1=0}$

$\mathtt{2y(y-1)\:\:+1\:(y-1) =0}$

$\mathtt{(y-1) \:\:(2y+1)=0}$

$\mathtt{y-1=0\:\:\:or\:\:2y+1=0}$

$\mathtt{y=1\:\:or\:\:2y=-1}$

$\mathtt{y=1\:\:or\:\:y=\dfrac{-1}{2}}$

Answer 2

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