The value of x2 - 6x + 13 can never be less than what ?
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x2−6x=13. x2−6x−13=0. Multiply the equation by 4. 4(x2−6x−13=0). 4x2−24x−52=0. (2x)2− 24x−52=0. (2x)2−2×(2x)×(6)+62−62−52=0.
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Heyy mate ❤✌✌❤
Here's your Answer....
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=> x^2 - 6x + 13.
Adding and subtracting 9 in this equation.
=> x^2 - 6x +9 - 9 + 13.
=> (x - 3)^2 + 4
Since, (X-3)^2 is always greater than 0.
Therefore, (x-3)^2 +4 will always be greater than 4.
✔✔✔
Here's your Answer....
⤵️⤵️⤵️⤵️⤵️⤵️
=> x^2 - 6x + 13.
Adding and subtracting 9 in this equation.
=> x^2 - 6x +9 - 9 + 13.
=> (x - 3)^2 + 4
Since, (X-3)^2 is always greater than 0.
Therefore, (x-3)^2 +4 will always be greater than 4.
✔✔✔
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