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Question: How do you solve \({x^2} - 12x + 35 \leqslant 0\)?...

How do you solve x212x+350{x^2} - 12x + 35 \leqslant 0?

Explanation

Solution

First of all, make the factors of the given expression using the method of splitting the middle term. Then, just use the fact that if a.b0a.b \leqslant 0, then either a0a \leqslant 0 and b>0b > 0 or a>0a > 0 and b0b \leqslant 0.

Complete step-by-step solution:
We are given that we are required to solve x212x+350{x^2} - 12x + 35 \leqslant 0.
Let us assume that f(x)=x212x+35f(x) = {x^2} - 12x + 35.
We will use the method of “splitting the middle term to factorize this function.
Therefore, we can write the given equation as f(x)=x25x7x+35f(x) = {x^2} - 5x - 7x + 35
Taking x common from the first two terms in the function’s expression, we will then obtain the following expression:-
f(x)=x(x5)7x+35\Rightarrow f(x) = x\left( {x - 5} \right) - 7x + 35
Taking – 7 common from the last two terms in the function’s expression, we will then obtain the following expression:-
f(x)=x(x5)7(x5)\Rightarrow f(x) = x\left( {x - 5} \right) - 7\left( {x - 5} \right)
Taking (x – 5) common from the last two terms in the function’s expression, we will then obtain the following expression:-
f(x)=(x5)(x7)\Rightarrow f(x) = \left( {x - 5} \right)\left( {x - 7} \right)
Now, putting this in the expression given to us, we will then obtain the following equation with us:-
(x5)(x7)0\Rightarrow \left( {x - 5} \right)\left( {x - 7} \right) \leqslant 0
Now, since we know that if a.b0a.b \leqslant 0, then either a0a \leqslant 0 and b0b \geqslant 0 or a0a \geqslant 0 and b0b \leqslant 0.
Replacing a by (x – 5) and b by (x – 7), we will then obtain the following expressions:-
Either x50x - 5 \leqslant 0 and x70x - 7 \geqslant 0 or x50x - 5 \geqslant 0 and x70x - 7 \leqslant 0 .
Either x5x \leqslant 5 and x7x \geqslant 7 or x5x \geqslant 5 and x7x \leqslant 7 .
x5\Rightarrow x \leqslant 5 and x7x \geqslant 7 are not possible together.
Therefore, x5x \geqslant 5 and x7x \leqslant 7 which implies that x[5,7]x \in \left[ {5,7} \right].

Note: The students must notice that we have an alternate way of factoring the quadratic equation involved in it as well. The alternate way is as follows:-
The given equation is x212x+35{x^2} - 12x + 35.
Let us equate the given equation to 0 for once so that we can find its roots easily.
So, the equation becomes x212x+35=0{x^2} - 12x + 35 = 0.
Using the quadratic formula given by if the equation is given by ax2+bx+c=0a{x^2} + bx + c = 0, its roots are given by the following equation:-
x=b±b24ac2a\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Thus, we have the roots of x212x+35{x^2} - 12x + 35 given by:
x=(12)±(12)24×352\Rightarrow x = \dfrac{{ - ( - 12) \pm \sqrt {{{( - 12)}^2} - 4 \times 35} }}{2}
Simplifying the calculations in the square root in the numerator of the right hand side, we will then obtain the following equation with us:-
x=12±1441402\Rightarrow x = \dfrac{{12 \pm \sqrt {144 - 140} }}{2}
Simplifying the calculations in the square root in the numerator of the right hand side further, we will then obtain the following equation with us:-
x=12±22\Rightarrow x = \dfrac{{12 \pm 2}}{2}
Hence, the roots are 5 and 7.
Thus, we have the required factors and we can proceed as we did in the solution.