Question

the quadratic equation whose one of the root is (3-√5)
a. x*2-6x+4=0
b.3x*2+5x+2=0
c. x*2-2x+7=0
d. 2x*2+3x+5=0​

Answer 1

Answer:

(3-√5) is the root of equation ${x}^{2} - 6x + 4 = 0$.So option a is the correct answer.

Step-by-step explanation:

Given root is $(3 - \sqrt{5} )$

Now we want find which equation has the given root.

For solving this question,we need to put value $(3 - \sqrt{5} )$instead of x.

i.e x=$(3 - \sqrt{5} )$

For the first equation ${x}^{2} -6x+4=0....(1)$

We are putting $x = (3 - \sqrt{5} )$

in the polynomial ${x}^{2} -6x+4$

We get,

${(3 - \sqrt{5}) }^{2} - 6(3 - \sqrt{5}) + 4$

$9 - 6 \sqrt{5} + 5 - 18 + 6 \sqrt{5} + 4 = 0$

So equation $(3 - \sqrt{5} )$is the root of equation (1),

Again for equation

$3 {x}^{2} +5x+2=0.....(2)$

We are putting $x = (3 - \sqrt{5} )$

in the polynomial $3 {x}^{2} +5x+2$

$3 {(3 - \sqrt{5}) }^{2} + 5(3 - \sqrt{5}) + 2 = 3(27 - 6 \sqrt{5} + 5) + 15 - 5 \sqrt{5} + 2 = 81 - 18 \sqrt{5} + 15 + 15 - 5 \sqrt{5} + 2$≠0

So $(3 - \sqrt{5} )$is the root of equation (2).

For equation ${x}^{2} -2x+7 = 0....(3)$

We are putting $x = 3 - \sqrt{5}$

in the polynomial ${x}^{2} -2x+7$

We get,${(3 - \sqrt{5}) }^{2} - 2(3 - \sqrt{5}) + 7 = 9 - 6 \sqrt{5} + 5 - 6 + 2 \sqrt{5} + 7$≠0

So $(3 - \sqrt{5} )$is not root of equation (3)

For equation $2 {x}^{2} +3x+5=0....(4)$

We are putting $x = 3 - \sqrt{5}$

in the polynomial $2 {x}^{2} +3x+5$

We get,$2 {(3 - \sqrt{5} )}^{2} + 3 \times (3 - \sqrt{5} ) + 5$

$2 \times (9 - 6 \sqrt{5} + 5) + 9 - 3 \sqrt{5} + 5 = 18 - 12 \sqrt{5} + 10 + 9 - 3 \sqrt{5} + 5$≠0

So,$3 - \sqrt{5}$is not root of equation (4).

Hence $(3 - \sqrt{5} )$is the root of equation (1).