AnswLet f(x)=x

2

−7x+12=0 be the quadratic equation

Now x=3.x=4 are the solutions, if f(3)=0,f(4)=0

Therefore, f(3)=3

2

−7(3)+12

=9−21+12

=0 and

and f(4)=4

2

−7(4)+12

=16−28+12

=0

Therefore x=3,x=4 are the solutions of the given equation.er:

Answer:

(1) x² - 7x + 12 = 0

x = 3,

LHS = (3)² - 7(3) + 12

= 9 - 21 + 12

= 21 - 21

= 0 = RHS

x = 3 is the solution

x = 4,

LHS = (4)² - 7(4) + 12

= 16 - 28 + 12

= 28 - 28

= 0 = RHS

x = 4 is the solution

[tex](2) {x}^{2} - \sqrt{2} x - 4 = 0[/tex]

[tex]at \: x = - \sqrt{2} [/tex]

[tex]LHS = {( - \sqrt{2}) }^{2} - \sqrt{2} ( - \sqrt{2} ) - 4[/tex]

[tex] = 2 + 2 - 4[/tex]

[tex] = 4 - 4[/tex]

[tex] = 0 = RHS[/tex]

[tex]x = - \sqrt{2} \: is \: the \:solution[/tex]

[tex]at \: x = 2 \sqrt{2} [/tex]

[tex]LHS = {( 2 \sqrt{2}) }^{2} - \sqrt{2} ( 2 \sqrt{2} ) - 4[/tex]

[tex] = 4(2) - 2(2) - 4[/tex]

= 8 - 4 - 4

= 8 - 8

= 0 = RHS

[tex]x = 2\sqrt{2} \: is \: the \:solution[/tex]