Question

if x is real, find the least value of x2-6x+10

Answer 1

Please see the attachment

Answer 2

Answer:

For x being real, Least value of $x^2 - 6x + 10 = 1$

Step-by-step explanation:

For x to be real.

Given: Quadratic Equation $x^2 -6x + 10 = 0$

Calculation:

We will find value of x by finding roots of equation.

Using Shri Dharacharya Formula:

$x = \frac{-b \frac{+}{-} \sqrt{ (b^2 -4ac)}}{2a}\\ a = 1\\b = -6\\c = 10$

Substituting values in formula and calculate for x.

$x = \frac{-(-6)\frac{+}{-}\sqrt{((-6)^2-(4*1*10))} }{2*1}$

Calculate for x

$x = \frac{6\frac{+}{-} \sqrt{36 - 40} }{2}\\ x = \frac{6\frac{+}{-} \sqrt{-4} }{2}$

Now value inside root is less than 0, then this term will be imaginary term

$x = \frac{6\frac{+}{-}\sqrt{4i^2} }{2}\\ x = \frac{3\frac{+}{-}2i }{1}\\ x = 3 \frac{+}{-} 2i$

for x to be real, $x > 0$

It means that,

We have $(x - 3 - 2i)(x -3 + 2i) > 0\\(x-3-2i) > 0\\or\\(x-3+2i) > 0$

It means x would be last real point

Hence least value of equation be:

$3^2 - 6*3 + 10 = 9 -18 + 10 = 19 - 18 = 1$

Least value is 1