Question
Question: Solve the quadratic equation \[{x^2} + 12x + 35 = 0\]....
Solve the quadratic equation x2+12x+35=0.
Solution
We have a polynomial of degree 2 and it is called a quadratic equation. We can solve this using the factorization method or by the quadratic formula. Let us solve this using the quadratic formula. That is we use the formula x=2a−b±b2−4ac. The quadratic formula is also called Sridhar’s formula.
Complete step-by-step solution:
Given, x2+12x+35=0.
Now consider the equation x2+12x+35=0. The degree of this equation is two hence we will have two factors.
On comparing the given equation with the standard quadratic equation ax2+bx+c=0. We have a=1, b=12 and c=35.
Now let us take the quadratic formula,
x=2a−b±b2−4ac.
Substituting the values we have,
⇒x=2(1)−12±122−4(1)(35)
⇒x=2−12±144−140
⇒x=2−12±4
We know that 4 is a perfect square,
⇒x=2−12±2
Thus we have two roots,
⇒x=2−12+2 and x=2−12−2
⇒x=2−10 and x=2−14
⇒x=−5 and x=−7. This is the required solution.
Additional information:
The highest exponent of the polynomial in a polynomial equation is called its degree. A polynomial equation has exactly as many roots as its degree. In various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors/solution/zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis, that is the roots are simply the x-intercepts.
Note: The above equation can be solved using the factorization method. That is for factorization, the standard equation is rewritten as ax2+b1x+b2x+c=0 such thatb1×b2=ac andb1+b2=b.
Here we can say that b1=7 and b2=5. Because b1×b2=35 (a×c) and b1+b2=12(b). If we split this and if we solve this we will have the same answer as above.