Question

x^2 + 7x -13 = (x+a)^2 + b

Find value of a and b by completing square​

Answer 1

How do I write x^2 - 7x + 5 in the form (x- a) ^2 - b?

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How do I write x^2 - 7x + 5 in the form (x- a) ^2 - b?

This often referred to by US teachers as “completing the square”. The key is you want to create a new term that is to equal to the square of half the linear term’s coefficient.

The linear term’s coefficient is -7. Half of -7 is -3.5. -3.5 squared is 12.25.

But in order to maintain the same value of my expression if I add 12.25 to the expression I must also subtract 12.25 from the expression.

Then I rewrite the equation as a perfect square binomial with an additional constant.

x² - 7x + 5 : add and subtract 12.25 to the expression

x² - 7x +(12.25) + 5 - (12.25) : simplify the last two terms

(x² - 7x + 12.25) - 7.25 : Rewrite the first three terms as a perfect square binomial, it will be (x - half the linear term)²

(x - 3.5)² - 7.25

a = 3.5 b = -7.25, as a result the vertex is at (3.5, -7.25)

Answer 2

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

$\sf {x}^{2} + 7x - 13 = {(x + a)}^{2} + b$

Completing the square method in which we remove the coefficient of x square by dividing the whole Equation with it but here coefficient is 1 so , no need to divide.

$\sf= > {\bigg(x - \dfrac{7}{2}\bigg )}^{2} = {\bigg(x + a\bigg)}^{2} + b$

$\sf= > {\bigg(x - \dfrac{7}{2}\bigg) }^{2} - 13 - {\bigg( \dfrac{7}{2} \bigg)}^{2} = {(x + a)}^{2} + b$

$\sf= > {\bigg(x - \dfrac{7}{2}\bigg) }^{2} - 13 - \dfrac{49}{4} = {\bigg(x + a\bigg)}^{2} + b$

$\sf = > {\bigg(x - \dfrac{7}{2}\bigg) }^{2} - \dfrac{52 - 49}{4} = {\bigg(x + a\bigg)}^{2} + b$

$\sf= > {\bigg(x - \dfrac{7}{2} \bigg)}^{2} - \dfrac{3}{4} = {\bigg(x + a\bigg)}^{2} + b.....(1)$

comparing with constant term and variable term separately.

$\sf = > x - \dfrac{7}{2} = x + a$

$\bold{a = - \dfrac{7}{2} }$

$\bold{ - \dfrac{3}{4} = b}$

Now check from Equation (1)

$\sf= > {\bigg(x + a\bigg)}^{2} + b$

$\sf= > {\bigg(x - \dfrac{7}{2}\bigg )}^{2} - \dfrac{3}{4}$