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Question: How do you find the discriminant and how many solutions does \(x^{2} +4x+3=4\) have?...

How do you find the discriminant and how many solutions does x2+4x+3=4x^{2} +4x+3=4 have?

Explanation

Solution

This type of questions is related to quadratic equations, we can solve this kind of questions with the help of formula for discriminant is D=d2=b2−4acD=d^{2} =b^{2} -4ac , where a is the coefficient of x2x^{2} , b is the xx-coefficient and c is the constant and for finding how many solutions does the given equation has, the formula is x=−b2a±d2ax=-\dfrac{b}{2a} \pm \dfrac{d}{2a} . And from this question we get, the a, b and c values and directly we are going to substitute into the formulae.

Complete step-by-step solution:
In the given question, we have the quadratic equation as x2+4x+3=4....................(i)x^{2} +4x+3=4....................(i)
We know that the general form of quadratic equation is ax2+bx+c=0..................(ii)ax^{2} +bx+c=0..................(ii)
So, now we are going to convert the equation(i) into the general form
Equation (i) becomes
x2+4x+3−4=0x^{2} +4x+3-4=0
x2+4x−1=0......................(iii)x^{2} +4x-1=0......................(iii)
Now compare the equation (iii) with equation (ii)
We get,
a=1;
b=4;
c=-1;
In quadratic equation, for finding the Discriminant
D=d2=b2−4acD=d^{2} =b^{2} -4ac
d2=b2−4ac.....................(iv)d^{2} =b^{2} -4ac.....................(iv)
Now substituting the values of a, b and c into equation(iv)
d2=42−4(1)(−1)d^{2} =4^{2} -4(1)(-1)
⇒d2=16+4\Rightarrow d^{2}= 16+4
⇒d=20\Rightarrow d= \sqrt{20}
d=±25d=\pm 2\sqrt{5}
We got the discriminant as d=±25d=\pm 2\sqrt{5}
For finding how many solutions does given equation have,
x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{b^{2} -4ac} }{2a}
x=−b2a±d2a(d=b2−4ac).....................(v)x=-\dfrac{b}{2a} \pm \dfrac{d}{2a} (d=\sqrt{b^{2} -4ac} ) .....................(v)
Now substitute again the values in equation(v)
x=−42(1)±252(1)x=\dfrac{-4}{2(1)} \pm \dfrac{2\sqrt{5} }{2(1)}
Then we finally got, x=−2±5x=-2\pm \sqrt{5}
Now x has two solutions and are real roots,
i.e.,x1=−2+5x_{1} =-2+\sqrt{5}
x2=−2−5x_{2} =-2- \sqrt{5}
Where x1x_{1} and x2x_{2} represents the roots of a quadratic equation (or) solution of the quadratic equation.

Note: While solving this kind of question, the common mistake done by the students is taking the square root before isolating the variable. Also, forgetting to put the plus or minus sign before the square root sign. Not understanding how to use the property of the square root. And a major mistake done by the students is cancelling incorrectly and also not knowing how to solve the radicals.