How do you factor \[8{x^2} + 10x - 3 = 0\]?
Answer
620.4k+ 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. If factors are difficult to find then we use Sridhar’s formula to find the roots. That is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Complete step-by-step solution:
Given, \[8{x^2} + 10x - 3 = 0\]
Since the degree of the equation is 2, we have 2 factors or two roots.
On comparing the given equation with the standard quadratic equation\[a{x^2} + bx + c = 0\], we have\[a = 8\], \[b = 10\] and \[c = - 3\].
The standard form of the factorization of quadratic equation is \[a{x^2} + {b_1}x + {b_2}x + c = 0\], which satisfies the condition \[{b_1} \times {b_2} = a \times c\] and \[{b_1} + {b_2} = b\].
We can see that we cannot split the middle term which satisfy above condition,
Now we use quadratic formula or Sridhar’s formula,
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Substituting we have,
\[ \Rightarrow x = \dfrac{{ - 10 \pm \sqrt {{{\left( {10} \right)}^2} - 4(8)( - 3)} }}{{2(8)}}\]
\[ = \dfrac{{ - 10 \pm \sqrt {100 - 4( - 24)} }}{{16}}\]
\[ = \dfrac{{ - 10 \pm \sqrt {100 + 96} }}{{16}}\]
\[ = \dfrac{{ - 10 \pm \sqrt {196} }}{{16}}\]
We know that 196 is a perfect square,
\[ = \dfrac{{ - 10 \pm 14}}{{16}}\]
Thus we have two roots,
\[ \Rightarrow x = \dfrac{{ - 10 + 14}}{{16}}\] and \[x = \dfrac{{ - 10 - 14}}{{16}}\]
\[ \Rightarrow x = \dfrac{4}{{16}}\] and \[x = \dfrac{{ - 24}}{{16}}\]
\[ \Rightarrow x = \dfrac{1}{4}\] and \[x = \dfrac{{ - 3}}{2}\]. This is the required answer.
Note: Since the degree of the polynomial is two hence we have 2 roots. In above we are unable to expand the middle term of the given equation into a sum of two numbers hence we use quadratic formula to solve the given problem. Quadratic formula and Sridhar’s formula are both the same. We know that the product of two negative numbers gives us a positive number. Also keep in mind when shifting values from one side of the equation to another side of the equation, always change sign from positive to negative and vice-versa.
Complete step-by-step solution:
Given, \[8{x^2} + 10x - 3 = 0\]
Since the degree of the equation is 2, we have 2 factors or two roots.
On comparing the given equation with the standard quadratic equation\[a{x^2} + bx + c = 0\], we have\[a = 8\], \[b = 10\] and \[c = - 3\].
The standard form of the factorization of quadratic equation is \[a{x^2} + {b_1}x + {b_2}x + c = 0\], which satisfies the condition \[{b_1} \times {b_2} = a \times c\] and \[{b_1} + {b_2} = b\].
We can see that we cannot split the middle term which satisfy above condition,
Now we use quadratic formula or Sridhar’s formula,
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Substituting we have,
\[ \Rightarrow x = \dfrac{{ - 10 \pm \sqrt {{{\left( {10} \right)}^2} - 4(8)( - 3)} }}{{2(8)}}\]
\[ = \dfrac{{ - 10 \pm \sqrt {100 - 4( - 24)} }}{{16}}\]
\[ = \dfrac{{ - 10 \pm \sqrt {100 + 96} }}{{16}}\]
\[ = \dfrac{{ - 10 \pm \sqrt {196} }}{{16}}\]
We know that 196 is a perfect square,
\[ = \dfrac{{ - 10 \pm 14}}{{16}}\]
Thus we have two roots,
\[ \Rightarrow x = \dfrac{{ - 10 + 14}}{{16}}\] and \[x = \dfrac{{ - 10 - 14}}{{16}}\]
\[ \Rightarrow x = \dfrac{4}{{16}}\] and \[x = \dfrac{{ - 24}}{{16}}\]
\[ \Rightarrow x = \dfrac{1}{4}\] and \[x = \dfrac{{ - 3}}{2}\]. This is the required answer.
Note: Since the degree of the polynomial is two hence we have 2 roots. In above we are unable to expand the middle term of the given equation into a sum of two numbers hence we use quadratic formula to solve the given problem. Quadratic formula and Sridhar’s formula are both the same. We know that the product of two negative numbers gives us a positive number. Also keep in mind when shifting values from one side of the equation to another side of the equation, always change sign from positive to negative and vice-versa.
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