Question

(11+x)×(5+1)=(3+x)×(15+x) find x​

Answer 1

I thing this is right but i thing this may be wrong

Answer 2

The value of x is -6+√57 or -6-√57

Explanation:

Given:

(11+x)×(5+1)=(3+x)×(15+x)

To find:

The value of x

Formula:

$\alpha =\frac{-b+\sqrt{b^{2}-4ac } }{2a}$

$\beta = \frac{-b-\sqrt{b^{2}-4ac } }{2a}$

Solution:

==> LHS = (11+x)×(5+1)

==> Solve LHS

==> LHS = (11+x)×(6)

==> LHS = 66+6x

==> RHS = (3+x)×(15+x)

==> Solve RHS

==> RHS = 3(15+x)+x(15+x)

==> RHS = 45+3x+15x+x²

==> RHS = x²+18x+45

==> LHS=RHS

==> 66+6x = x²+18x+45

==> x²+18x+45 -66-6x=0

==> x²+12x+45-66=0

==> x²+12x-21=0

==> Solve the quadratic equation x²+12x-21=0

==> a = coefficient of x²

==> b = coefficient of x

==> c = constant

==> a = 1

==> b = 12

==> c = -21

==> Apply the values in the formula

==>  $\alpha =\frac{-b+\sqrt{b^{2}-4ac } }{2a}$

==> $\alpha =\frac{-12+\sqrt{12^{2}-4(1)(-21) } }{2(1)}$

==> $\alpha =\frac{-12+\sqrt{144-4(-21) } }{2}$

==> $\alpha =\frac{-12+\sqrt{144+84 } }{2}$

==> $\alpha =\frac{-12+\sqrt{ 228} }{2}$

==> $\alpha =\frac{-12+\sqrt{ 2\times2\times3\times19} }{2}$

==> $\alpha =\frac{-12+2\sqrt{ 3\times19} }{2}$

==> $\alpha =\frac{2(-6+\sqrt{ 57} )}{2}$

==> α = -6+√57

==>  $\beta =\frac{-b-\sqrt{b^{2}-4ac } }{2a}$

==> $\beta =\frac{-12-\sqrt{12^{2}-4(1)(-21) } }{2(1)}$

==> $\beta =\frac{-12-\sqrt{144-4(-21) } }{2}$

==> $\beta =\frac{-12-\sqrt{144+84 } }{2}$

==> $\beta =\frac{-12-\sqrt{ 228} }{2}$

==> $\beta =\frac{-12-\sqrt{ 2\times2\times3\times19} }{2}$

==> $\beta =\frac{-12-2\sqrt{ 3\times19} }{2}$

==> $\beta =\frac{2(-6-\sqrt{ 57} )}{2}$

==> β = -6-√57

The roots of the equation x²+12x-21=0 are  α = -6+√57 and β = -6-√57

The value of x is -6+√57 or -6-√57