Solve the quadratics, use any method: a) y=x2+16x−5 b) x2−6x−99=−9 Identify the vertex, the y-intercept.

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Solve the quadratics, use any method:  a) y=x2+16x−5  b) x2−6x−99=−9  Identify the vertex, the y-intercept.

Shawn Kim · Accepted Answer

Step 1  Given: quadratic equation  x2−6x−99=−9  Step 2  Explanation  Let the quadratic of the form  ax2+bx+c=0  Then the value of x can be using the formula  x=−b±b2−4ac2a  a is coefficient of x2  b is coefficient of x  and c is term  and b2−4ac is discriminant △  if △&amp;gt;0 the roots are real and disfinct  △=0 the roots are real and equal  △&amp;lt;0 the roots are complex i.e. not-real  Step 3  If we rewrite the equation  x2−6x−99=−9  ⇒x2−6x−99+9=0  ⇒x2−6x−90=0  a=1  b=−6  c=−90  So finding the value of x by all the values  x=−b±b2−4ac2a  =−(−6)±(−6)2−4×1×(−40)2×1  =6±36+3602  =6±3962  =6±6112  =3±311  Hence, the values of the zeroes of equation x2−6x−90=0 are 3+311, 3−311

lenkiklisg7 · Accepted Answer

Step 1  a) y=x2+16x−5  Swap sides so that all variable terms are on the left hand side.  x2+16x−5=y  Subtract y from both sides.  All equations of the form ax2+bx+c=0 can be solved using the quadratic formula:−b±b2−4ac2a The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.  x2+16x−y−5=0  This equation is in standard form. Substitute 1 for a, 16 for b, and −5−y for c in the quadratic formula,  x=−16±162−4(−y−5)2  Square 16.  x=−16±256−4(−y−5)2  Multiply -4 times -5-y.  x=−16±256+4y+202  Add 256 to 20+4y.  x=−16±4y+2762  Take the square root of 276+4y.  x=−16±2y+692  Now solve the equation x=−16±2y+692 when ± is plus. Add −16 to 269+y  x=2y+69−162  Divide −16+269+y by 2  x=y+69−8  Now solve the equation x=−16±2y+692 when ± is minus. Subtract 269+y from -16  x=−2y+69−162  Divide −16−269+y by 2  x=−y+69−8  The equation is now solved  x=y+69−8  x=−y+69−8

Solve the quadratics, use any method: a) y=x^{2}+16x-5 b) x^{2}-6x-99=-9 Identify the vertex,

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