How do you solve \[8{m^2} - 2m = 7\] using the quadratic formula?
Answer
601.5k+ views
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. Here we have ‘m’ instead of ‘x’ as variable. Reaming we follow the same procedure. Since they are asking to solve using quadratic formula we solve this using quadratic formula that is\[m = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Complete step-by-step solution:
Given, \[8{m^2} - 2m = 7\]
Rearranging the equation we have,
\[8{m^2} - 2m - 7 = 0\]
Since the degree of the equation is 2, we have 2 roots.
On comparing the given equation with the standard quadratic equation\[a{m^2} + bm + c = 0\], we have\[a = 8\], \[b = - 2\] and \[c = - 7\].
We have the quadratic formula,
\[m = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Substituting we have,
\[ \Rightarrow m = \dfrac{{ - ( - 2) \pm \sqrt {{{( - 2)}^2} - 4(8)( - 7)} }}{{2(8)}}\]
\[ \Rightarrow \dfrac{{4 \pm \sqrt {4 - 4( - 56)} }}{{16}}\]
\[ \Rightarrow \dfrac{{4 \pm \sqrt {4 + 224} }}{{16}}\]
\[ \Rightarrow \dfrac{{4 \pm \sqrt {228} }}{{16}}\]
But we can write \[228 = 4 \times 57\],
\[ \Rightarrow \dfrac{{4 \pm \sqrt {4 \times 57} }}{{16}}\]
\[ \Rightarrow \dfrac{{4 \pm 2\sqrt {57} }}{{16}}\]
Taking 2 common we have,
\[ \Rightarrow \dfrac{{2(2 \pm \sqrt {57} )}}{{16}}\]
Cancelling we have,
\[ \Rightarrow \dfrac{{2 \pm \sqrt {57} }}{8}\]
Thus we have two roots
\[ \Rightarrow m = \dfrac{{2 + \sqrt {57} }}{8}\] and \[m = \dfrac{{2 - \sqrt {57} }}{8}\].
Hence, the solutions of \[8{m^2} - 2m = 7\] are \[m = \dfrac{{2 + \sqrt {57} }}{8}\] and \[m = \dfrac{{2 - \sqrt {57} }}{8}\].
Note: 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.
Complete step-by-step solution:
Given, \[8{m^2} - 2m = 7\]
Rearranging the equation we have,
\[8{m^2} - 2m - 7 = 0\]
Since the degree of the equation is 2, we have 2 roots.
On comparing the given equation with the standard quadratic equation\[a{m^2} + bm + c = 0\], we have\[a = 8\], \[b = - 2\] and \[c = - 7\].
We have the quadratic formula,
\[m = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Substituting we have,
\[ \Rightarrow m = \dfrac{{ - ( - 2) \pm \sqrt {{{( - 2)}^2} - 4(8)( - 7)} }}{{2(8)}}\]
\[ \Rightarrow \dfrac{{4 \pm \sqrt {4 - 4( - 56)} }}{{16}}\]
\[ \Rightarrow \dfrac{{4 \pm \sqrt {4 + 224} }}{{16}}\]
\[ \Rightarrow \dfrac{{4 \pm \sqrt {228} }}{{16}}\]
But we can write \[228 = 4 \times 57\],
\[ \Rightarrow \dfrac{{4 \pm \sqrt {4 \times 57} }}{{16}}\]
\[ \Rightarrow \dfrac{{4 \pm 2\sqrt {57} }}{{16}}\]
Taking 2 common we have,
\[ \Rightarrow \dfrac{{2(2 \pm \sqrt {57} )}}{{16}}\]
Cancelling we have,
\[ \Rightarrow \dfrac{{2 \pm \sqrt {57} }}{8}\]
Thus we have two roots
\[ \Rightarrow m = \dfrac{{2 + \sqrt {57} }}{8}\] and \[m = \dfrac{{2 - \sqrt {57} }}{8}\].
Hence, the solutions of \[8{m^2} - 2m = 7\] are \[m = \dfrac{{2 + \sqrt {57} }}{8}\] and \[m = \dfrac{{2 - \sqrt {57} }}{8}\].
Note: 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.
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