Question

- The number that satisfies the
statement "The sum of a number
and twice its square is 105", is​

Answer 1

QUADRATIC EQUATIONS :

Given statement : " The sum of a number and twice its square is 105."

Let the number be x.

Now, twice of this number's square = $\mathsf{2{x}^{2}\:=\:105}$

Now, as per the question's statement :

Equation formed : $\boxed{\mathsf{x\:+\:2{x}^{2}\:=\:105}}$

$\mathsf{2{x}^{2}\:+\:x \:=\:105}$

$\mathsf{2{x}^{2}\:+\:x\:-105\:=\:0}$

By using splitting the middle term method,

$\mathsf{2{x}^{2}\:+\:15x\:-14x\:-105\:=\:0}$

$\mathsf{x(\:2x\:+\:15\:) \:-7(\:2x\:+\:15\:)\:=\:0}$

$\mathsf{(\:x\:-\:7\:)(\:2x\:+\:15\:)\:=\:0}$

$\mathsf{(\:x\:-\:7\:)\:=\:0,(\:2x\:+\:15\:)\:=\:0}$

$\boxed{\mathsf{x\:=\:7,\:{\dfrac{-15}{2}}}}$

VERIFICATION :

For x = 7,

Putting value of 'x' in equation,

$\mathsf{L. H. S. \:=\:2{7}^{2}\:+\:7 }$

$\mathsf{L. H. S. \:=\:2{\times{49}\:+\:7 }}$

$\mathsf{L. H. S. \:=\:105}$

L.H.S. = R. H. S.

For $\mathsf{x\:=\:{\dfrac{-15}{2}}}$,

Putting this value of 'x' in equation,

$\mathsf{L. H. S. \:=\:2({\dfrac{-15^2}{2^2})\:-\:{\dfrac{15}{2}}}}$

$\mathsf{L. H. S. \:=\:{\dfrac{225}{2}\:-\:{\dfrac{15}{2}}}}$

$\mathsf{L. H. S. \:=\:{\dfrac{210}{2}}}$

$\mathsf{L. H. S. \:=\:105}$

L. H. S. = R. H. S.

Hence, Both values of 'x' are correct.

Answer 2

Let the unknown number be denoted by the variable x.

Forming an equation from the given word equation:

x + 2x² = 105.

The equation formed above can be rearranged as 2x² + x - 105 = 0, and is a quadratic equation of the form ax² + bx + c, where a = 2, b = 1 and c = -105.

Using the quadratic formula to find the solutions of the equation:

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x = \frac{ - 1 \pm \sqrt{1 - 4(2)( - 105)} }{2(2)} \\ \\ x = \frac{ - 1 \pm \sqrt{1 + 840} }{4} \\ \\ x = \frac{ - 1 \pm \sqrt{841} }{4} \\ \\ x = \frac{ - 1 \pm29}{4} \\ \\ x = \frac{ - 1 - 29}{4} \\ \\ = \frac{ - 30}{4} \\ \\ = - 7.5 \: \: (or) \\ \\ x = \frac{ - 1 + 29}{4} = \frac{28}{4} = 7$

Therefore, the number can be -7.5 or 7.