Question

If 4 is the root of the quadratic equation \[{x^2} + ax - 8 = 0\] then a =…..
A) 2
B) 4
C) -2
D) -4

Answer 1

Hint:
Given equation is a quadratic equation of the form \[a{x^2} + bx + c = 0\]. Roots of this equation are obtained with the help of \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. But one of the roots is already given to us so we can equate one of the roots with 4 and find the value of a. Let’s solve it!

Complete step by step solution:
Given equation is \[{x^2} + ax - 8 = 0\]
Now comparing with the general quadratic equation we get a=1, b=a and c=-8.
Thus to find the roots let’s use \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
So one of the roots is 4.
\[ \Rightarrow \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} = 4\]
Putting the values,
\[ \Rightarrow \dfrac{{ - a + \sqrt {{a^2} - 4 \times 1 \times \left( { - 8} \right)} }}{{2 \times 1}} = 4\]
On cross multiplying,
\[ \Rightarrow - a + \sqrt {{a^2} - 4 \times 1 \times \left( { - 8} \right)} = 4 \times 2\]
\[ \Rightarrow - a + \sqrt {{a^2} - 4 \times 1 \times \left( { - 8} \right)} = 8\]
On solving the square root we get,
\[ \Rightarrow - a + \sqrt {{a^2} + 32} = 8\]
Shift a on right side
\[ \Rightarrow \sqrt {{a^2} + 32} = 8 + a\]
On squaring both sides we get,
\[ \Rightarrow {a^2} + 32 = {\left( {8 + a} \right)^2}\]
Performing the \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]
\[ \Rightarrow {a^2} + 32 = 64 + 2 \times 8 \times a + {a^2}\]
Cancelling \[{a^2}\] from both sides
\[ \Rightarrow 32 = 64 + 16a\]
\[ \Rightarrow 32 - 64 = 16a\]
Subtract the terms
\[ \Rightarrow - 32 = 16a\]
Divide 32 by 16
\[ \Rightarrow - \dfrac{{32}}{{16}} = a\]
\[ \Rightarrow - 2 = a\]
\[ \Rightarrow a = - 2\] thus value of a is found.

Hence the correct option is C.

Note:
Students can also use other roots and equate it with 4. We will get the same answer. In this problem when we solve the equated equation carefully look at the signs because a minus or plus sign can change the option.