If x be real, then minimum value of {x^2}

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4-2.Quadratic Equations and Inequations

easy

If $x$ be real, then the minimum value of ${x^2} - 8x + 17$ is

$-1$

$0$

$1$

$2$

Solution

(c) ${x^2} – 8x + 17={x^2} – 8x + 16+1={(x-4)^2} + 1$
Since $x$ is real, so ${(x – 4)^2}$ is always positive and its least value is $0$ and so the minimum value of given expression is $1.$

Standard 11

Mathematics

Suppose $a, b, c$ are positive integers such that $2^a+4^b+8^c=328$. Then, $\frac{a+2 b+3 c}{a b c}$ is equal to

hard

(KVPY-2015)

The number of ordered pairs $(x, y)$ of real numbers that satisfy the simultaneous equations $x+y^2=x^2+y=12$ is

normal

(KVPY-2015)

The set of values of $x$ which satisfy $5x + 2 < 3x + 8$ and $\frac{{x + 2}}{{x – 1}} < 4,$ is

easy

If $\alpha ,\beta $ and $\gamma $ are the roots of ${x^3} + px + q = 0$, then the value of ${\alpha ^3} + {\beta ^3} + {\gamma ^3}$ is equal to

hard

If $x$ is real , the maximum value of $\frac{{3{x^2} + 9x + 17}}{{3{x^2} + 9x + 7}}$ is

hard

(AIEEE-2006)

If $x$ be real, then the minimum value of ${x^2} - 8x + 17$ is

Solution

Similar Questions

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