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Question

Question: How do you solve \(3{x^2} + 6x = 8\)?...

How do you solve 3x2+6x=83{x^2} + 6x = 8?

Explanation

Solution

We are given a quadratic algebraic expression of the form ax2+bx+ca{x^2} + bx + c it is a quadratic expression. And we have to find its factor. Here the first term is ax2a{x^2} its coefficient isaa. The middle term is bx - bx. Coefficient is bb. The last term is constant i.e. c - c. First we will equate the equation equal to zero then by using the quadratic formula we find the value of xx. Quadratic formula states:

Formula used:
x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Here bb= coefficient of xx, coefficient of  x2\;{x^2}and  c\;c= constant term. After substituting the values in the formula we will get two values ofxx.

Complete step by step answer:
Given a quadratic equation and we have to find the value of xx. The given quadratic equation is 3x2+6x=83{x^2} + 6x = 8 so first we will equate the equation equal to zero. For this subtract 88 from both sides we will get:
3x2+6x8=88\Rightarrow 3{x^2} + 6x - 8 = 8 - 8
3x2+6x8=0\Rightarrow 3{x^2} + 6x - 8 = 0

Using the quadratic formula we will get the value of x.
x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Here a=3,b=6,c=8a = 3,b = - 6,c = - 8. Substituting these values in the equation we will get the value of xx.
6±(6)24.(3×8)2×3\Rightarrow \dfrac{{6 \pm \sqrt {{{\left( { - 6} \right)}^2} - 4.(3 \times 8)} }}{{2 \times 3}}
Now we will raise 6 - 6 to the power of 22.
x=6±364.(3.8)2×3\Rightarrow x = \dfrac{{6 \pm \sqrt {36 - 4.\left( {3. - 8} \right)} }}{{2 \times 3}}
Now we will multiply 33 by 8 - 8 and open the bracket by multiplying it by4 - 4.
x=6±364.(24)2×3\Rightarrow x = \dfrac{{6 \pm \sqrt {36 - 4.\left( { - 24} \right)} }}{{2 \times 3}}
x=6±36+962×3\Rightarrow x = \dfrac{{6 \pm \sqrt {36 + 96} }}{{2 \times 3}}

Now on adding 3636 and   96\;96.
x=6±1322×3\Rightarrow x = \dfrac{{6 \pm \sqrt {132} }}{{2 \times 3}}
Now we will rewrite 132132as22×33{2^2} \times 33
x=6±22×332×3\Rightarrow x = \dfrac{{6 \pm \sqrt {{2^2} \times 33} }}{{2 \times 3}}
Now we will pull out the terms 22{2^2} under the radical. Also solve the denominator.
x=6±2336\Rightarrow x = \dfrac{{6 \pm 2\sqrt {33} }}{6}
On simplifying the expression we will get:
x=3±333\Rightarrow x = \dfrac{{3 \pm \sqrt {33} }}{3}
Converting these in decimal form we will get:
x=4.917;x=1.083x = 4.917; x = - 1.083
Hence the value of x is x=4.917;x=1.083x = 4.917;x = - 1.083.

Note: In such types of questions if we are given a quadratic equation then it can be easily solved by splitting the middle term. If the equation is a simple form but if the equation contains big coefficients or any such coefficients that cannot be solved by splitting the middle term. Then use the quadratic formula i.e.b±b24ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} by using this we can solve any complicated equation. But for small simple equation then we will use the splitting the middle term method for this also there are two types of equation form for ax2+bx+ca{x^2} + bx + c we will split the middle term in form of addition and for ax2+bxca{x^2} + bx - c. We will split the middle term in form subtraction. Remember this while solving by splitting the middle term.