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Question

Question: How do you solve \[2{{x}^{2}}-9x+7=0\]?...

How do you solve 2x29x+7=02{{x}^{2}}-9x+7=0?

Explanation

Solution

Use the middle term split method to factorize 2x29x+72{{x}^{2}}-9x+7 and write it as a product of two terms given as (xa)(xb)\left( x-a \right)\left( x-b \right) where ‘a’ and ‘b’ are called zeroes of the polynomial. Now, substitute each term equal to 0 and find the two values of x to get the answer.

Complete step-by-step solution:
Here, we have been provided with the quadratic equation 2x29x+7=02{{x}^{2}}-9x+7=0 and we are asked to solve it. That means we have to find the values of x.
Now, let us apply the middle term split method to factorize the given equation first. Let us assume 2x29x+7=f(x)2{{x}^{2}}-9x+7=f\left( x \right), so we have,
f(x)=2x29x+7\Rightarrow f\left( x \right)=2{{x}^{2}}-9x+7
According to the middle term split method we have to break -9x into two terms such that their sum is -9x and product is equal to the product of constant term (7) and 2x22{{x}^{2}}, i.e., 14x214{{x}^{2}}. So, breaking -9x into -7x and -2x, we get,
(i) (7x)+(2x)=9x\left( -7x \right)+\left( -2x \right)=-9x
(ii) (7x)×(2x)=14x2\left( -7x \right)\times \left( -2x \right)=14{{x}^{2}}
Therefore, both the conditions of the middle term split method are satisfied, so we can write the given expression as: -

& \Rightarrow f\left( x \right)=2{{x}^{2}}-7x-2x+7 \\\ & \Rightarrow f\left( x \right)=x\left( 2x-7 \right)-1\left( 2x-7 \right) \\\ & \Rightarrow f\left( x \right)=\left( 2x-7 \right)\left( x-1 \right) \\\ \end{aligned}$$ Now, we have $$f\left( x \right)=0$$ $$\Rightarrow \left( 2x-7 \right)\left( x-1 \right)=0$$ Substituting each term equal to 0, we get, $$\Rightarrow \left( 2x-7 \right)=0$$ or $$\left( x-1 \right)=0$$ $$\Rightarrow x=\dfrac{7}{2}$$ or x = 1 **Hence, the solutions of the given equations are: - x = 1 or $$x=\dfrac{7}{2}$$.** **Note:** One may note that here we have applied the middle term split method to get the values of x. You can also apply the discriminant method to get the answer. In that condition assume the coefficient of $${{x}^{2}}$$ as a, the coefficient of x as b and the constant term as c, and apply the formula: - $$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$ to solve for the values of x. There can be a third method also, known as completing the square method, to solve the question. Note that the discriminant formula is obtained from completing the square method.