Question

x^2+3x+1=(x-2)^2 class 10 ncert ex.4.1 question 1 6th sub division

Answer 1

Answer:

Step-by-step explanation:

That quadratic equation has 2 roots, which can be calculated using the quadratic formula: 3+5√2 and 3−5√2

And using the first root:

x2−3x+1=0⟹x2=3x−1=33+5√2−1=7+35√2

Then 1x2=27+35√=2(7−35√)49–45=7−35√2

Adding them together:

x2+1x2=7+35√2+7−35√2=7

(Note that you’d get the same result for each of the roots)

Answer 2

$\star\bold\red{Answer:-}$

${x}^{2} +3x +1 = {(x-2)}^{2}$

${x}^{2} +3x +1 = {x}^{2} +{2}^{2} -2×x×2$

${x}^{2} +3x +1 = {x}^{2} +4 -4x$

${x}^{2} - {x}^{2} +3x +4x +1 -4 = 0$

$7x-3 = 0$

$7x = 3$

$x = \frac{3}{7}$