Answer:
[tex] {x}^{2} + 7x + \frac{49}{4} = {(x + \frac{7}{2}) }^{2} [/tex]
Explanation:
[tex] {x}^{2} + 7x + a = {(x + b)}^{2} [/tex]
[tex] {x}^{2} + 7x + a = {x}^{2} + 2bx + {b}^{2} [/tex]
compare the x co-efficient
[tex] 7 = 2b[/tex]
[tex] b = \frac{7}{2} [/tex]
compare the constants
[tex]a = {b}^{2} [/tex]
[tex]a = {( \frac{7}{2}) }^{2} [/tex]
[tex]a = \frac{49}{4} [/tex]
HOPE IT HELPS....
The complete equation will be x^2+7x+49/4=(x+7/2)2
Given the quadratic function x^2 + 7x + ?
In order to complete the square using the completing the square method, we will add the square of the half of coefficient of x to both sides of the expression.
Coefficient of x = 7
Half of the coefficient = 7/2
Taking the square of the result = (7/2)² = 49/4
The constant that will complete the equation is 49/9. The equation becomes x^2 + 7x + (7/2)² = (x+7/2)²
Hence the complete equation will be x^2+7x+49/4=(x+7/2)2
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