Question

How do you solve \[7{x^2} + 10x = 2{x^2} + 155\] using any method?

Answer 1

Hint: A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. The quadratic formula is used when we fail to find the factors of the equation. We first express this in standard form. Then we apply the quadratic formula that is,\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step-by-step solution:
Given, \[7{x^2} + 10x = 2{x^2} + 155\]
Rearranging the equation we have,
\[7{x^2} + 10x - 2{x^2} - 155 = 0\]
\[5{x^2} + 10x - 155 = 0\]
Divide by 5 on both side of the equation we have,
\[{x^2} + 2x - 31 = 0\]
Since the degree of the equation is 2, we have 2 roots.
On comparing the given equation with the standard quadratic equation\[a{x^2} + bx + c = 0\], we have\[a = 1\], \[b = 2\] and \[c = - 31\].
We have the quadratic formula,
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Substituting we have,
\[ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4(1)( - 31)} }}{{2(1)}}\]
\[ = \dfrac{{ - 2 \pm \sqrt {4 + 124} }}{2}\]
\[ = \dfrac{{ - 2 \pm \sqrt {128} }}{2}\]
We can write \[128 = 64 \times 2\]
\[ = \dfrac{{ - 2 \pm \sqrt {64 \times 2} }}{2}\]
We know that 64 is a perfect square
\[ = \dfrac{{ - 2 \pm 8\sqrt 2 }}{2}\]
Taking 2 common we have,
\[ = \dfrac{{2( - 1 \pm 4\sqrt 2 )}}{2}\]
\[ = - 1 \pm 4\sqrt 2 \]
Thus we have two roots,
\[ \Rightarrow x = - 1 + 4\sqrt 2 \] and \[x = - 1 - 4\sqrt 2 \].
Hence, the solutions of \[7{x^2} + 10x = 2{x^2} + 155\] are \[x = - 1 + 4\sqrt 2 \] and \[x = - 1 - 4\sqrt 2 \].

Note: Quadratic formula and Sridhar’s formula are both the same. If a polynomial is of degree ‘n’ then we have ‘n’ roots. Here the degree of the polynomial is 2 hence we have 2 roots. In various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors/solution/zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts. The quadratic formula is also called Sridhar’s formula. Careful in the calculation part.